![22 Electric Field
Now flux in this surface is
dφ = E × dS1
Or
dφ =
1
4πε0εr
×
q2
~
r2
2
× π(~
r2
2 − ~
r2
2 cos2
θ2)
It gives
dφ =
1
4ε0εr
q2(1 − cos2
θ2)
If θ ≈ 0 then (1 − cos2
θ2) = 2(1 − cos θ2). Now
dφ =
1
2ε0εr
q2(1 − cos θ2)
It is clear that flux over the given surface is depend only on the charge and
angular position of point from the charge. Now we can take that
q2(1 − cos θ2) = K
Where K is a constant value. If there are n charge system enclosed by virtual
Gaussian surface then
i=n
X
i=1
qi(1 − cos θi) = K1
Or
i=n
X
i=1
qi cos θi = K2
cos θ for a point charge at placed (x1, y1) with respect point (x, y) is given by
cos θ =
x − x1
p
(x − x1)2 + (y − y1)2
Hence for n point charge system
i=n
X
i=1
qi
x − xi
p
(x − xi)2 + (y − yi)2
= K2
Now we can calculate all possible points for different K2 of electric field lines
for which above relation is satisfied. Now the path of these points is the trace
of electric field line. C program for generation of tracing points of electric
potential surface of two charge system at [1C, −2x, 0y] and [−1C, 2x, 0y] is
given below.](https://image.slidesharecdn.com/electricfield-211010144000/85/Principle-of-Electric-Field-Physics-by-Arun-Umrao-38-320.jpg)
![1.2. NUMERICAL COMPUTATION 23
✞
1 #include stdio.h
#include stdlib.h
3 #include math.h
#define bufsize 100
5 char r[bufsize ];
FILE *fp;
7 int n;
9 int main(int argc , char** argv) {
float x, y, k;
11 fp = fopen(a.txt, wb);
for (x = -5; x 5; x = x + 0.1) {
13 for (y = -5; y 5; y = y + 0.1) {
k = (x + 2) / pow((x + 2)*(x + 2) + y*y,
0.5)
15 - (x - 2) / pow((x - 2)*(x - 2) + y*y,
0.5);
if(k 0.2409 k 0.25001) {
17 n = sprintf (r, %2.2ft%2.3ft%2.3fn,
k, x, y);
fwrite(r, sizeof (char), n, fp);
19 }
}
21 }
fclose(fp);
23 return (EXIT_SUCCESS);
}
✌
✆
A tabulted data table is saved in the file ‘a.txt’.
Solved Problem 1.6 Find the tabulated data for electric field lines for charge
system 1C and −0.5C placed at (−1, 0) and (2, 0) respectively.
Solution
✞
#include stdio.h
2 #include stdlib.h
#include math.h
4 #define bufsize 100
char r[bufsize ];
6 FILE *fp;](https://image.slidesharecdn.com/electricfield-211010144000/85/Principle-of-Electric-Field-Physics-by-Arun-Umrao-39-320.jpg)
![1.2. NUMERICAL COMPUTATION 25
At a point P(x, y) electric potential is given by V = V1 + V2 and from
potential definitions:
V =
1
4πε0εr
q1
r1
+
q2
r2
It is clear that potential at a point is depend only on the sum of ratios of
charge to distance.
q1
r1
+
q2
r2
= K
Where K is a constant value. If there are n charges in a system then
i=n
X
i=1
qi
ri
= K1
Here
ri =
p
(x − xi)2 + (y − yi)2
Hence for n point charge system
i=n
X
i=1
qi
p
(x − xi)2 + (y − yi)2
= K1
Now we can calculate all possible points for different K1 of electric poten-
tial surfaces for which above relation is satisfied. Now the path of these
points is the trace of electric field line. C program for generation of tracing
points of electric potential surface of two charge system at [1C, −2x, 0y] and
[−1C, 2x, 0y] is given below.
✞
#include stdio.h
2 #include stdlib.h
#include math.h
4 #define bufsize 100
char r[bufsize ];
6 FILE *fp;
int n;
8
int main(int argc , char** argv) {
10 float x, y, k;
fp = fopen(a.txt, wb);
12 for (x = -5; x 5; x = x + 0.1) {
for (y = -5; y 5; y = y + 0.1) {](https://image.slidesharecdn.com/electricfield-211010144000/85/Principle-of-Electric-Field-Physics-by-Arun-Umrao-41-320.jpg)
![1.4. ELECTRIC FORCE 31
This is also known as relation between electric field and electric potential.
Work Done When Electric Dipole Is In Electric Field
With reference from figure (1.12), Force on electric dipole when it is placed
in external electric field is τ = p ~
E sin α. Work required to rotate it a small
angle dα is dW hence
dW = p ~
E sin α dα
on integrating above equation within limit α = θ1 to α = θ2 total work is
W = −p ~
E [cos α]θ2
θ1
~
E → ~
p
bcb θ
dθ
Figure 1.12: Work done in rotating electric dipole in an electric field.
Applying the limits
W = −p ~
E [cos θ2 − cos θ1]
W will be maximum when θ2 = 180◦
→ cos θ = −1 and θ1 = 0◦
→ cos θ = 1.
Now work done is
W = −p ~
E(−1 − 1) = 2p ~
E
Solved Problem 1.9 An electric dipole with dipole moment 4 × 10−9
C-m
is aligned at 30◦
with the direction of a uniform electric field of magnitude
5 × 104
N/C. Calculate the magnitude of the torque acting on the electric
dipole.
Solution Torque on a dipole is given by
τ = pE sin θ](https://image.slidesharecdn.com/electricfield-211010144000/85/Principle-of-Electric-Field-Physics-by-Arun-Umrao-47-320.jpg)
Principle of Electric Field - Physics - by Arun Umrao
- 1. 1 ELECTRIC FIELD A SHORT NOTES Arun Umrao https://sites.google.com/view/arunumrao DRAFT COPY - GPL LICENSING
- 2. 2 Electric Field Contents 1 Electric Field 3 1.1 Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1.1 Electric Field (Quantitatively Approach) . . . . . . . . 3 1.1.2 Electric Dipole . . . . . . . . . . . . . . . . . . . . . . 6 1.1.3 Electric Field Due To Electric Dipole . . . . . . . . . . 6 1.1.4 3-Dimensional Electric Field Problems . . . . . . . . . 16 1.2 Numerical Computation . . . . . . . . . . . . . . . . . . . . . 20 1.2.1 Electric Field . . . . . . . . . . . . . . . . . . . . . . . 20 1.2.2 Electric Potential . . . . . . . . . . . . . . . . . . . . . 24 1.3 Displacement Field . . . . . . . . . . . . . . . . . . . . . . . . 26 1.4 Electric Force . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 1.4.1 Electric Dipole in Electric Field . . . . . . . . . . . . . 27 1.4.2 Oil Drop Experiment . . . . . . . . . . . . . . . . . . . 32 1.5 Charge Density . . . . . . . . . . . . . . . . . . . . . . . . . . 34 1.6 Motion of Charged Particle in Electric Field . . . . . . . . . . 34 1.6.1 Motion of Charge in Uniform Electric Field . . . . . . 34 1.7 Relative Permittivity . . . . . . . . . . . . . . . . . . . . . . . 37 1.8 Electric Field Due To Charge Distribution . . . . . . . . . . . 37 1.8.1 Charged Rod At Axial Position . . . . . . . . . . . . . 38 1.8.2 Charged Rod On Equatorial Position . . . . . . . . . . 39 1.8.3 Charged Rod On Un-Symmetrical Position . . . . . . . 42 1.8.4 Charged Ring At Position On Its Axis . . . . . . . . . 45 1.8.5 Charged Disk On Its Axis . . . . . . . . . . . . . . . . 46 1.8.6 Cavity in a Non-Conducting Sphere . . . . . . . . . . . 49 1.9 Electric Force on Surface of Conductor . . . . . . . . . . . . . 51
- 3. 1.1. ELECTRIC FIELD 3 1Electric Field 1.1 Electric Field Electric field is the region around a charged body within which it can exert its electrostatic influence. Electric field is extends to infinity but it falls rapidly with distance. Electric field is a vector quantity. 1.1.1 Electric Field (Quantitatively Approach) A charge q2 is placed at finite distance from charge q1 then they experience an electrostatic force. Space around charge q1 where other charge q2 experience electrostatic force is called electric field of charge q1. Now the electric field of the charge q1 at the point of charge q2 is ~ E = Fq1q2 q2 (1.1) Where Fq1q2 is force between two charge bodies. Space around charge or charged body within which an experimental unit positive charge is attracted or repulsed by charge or charged body is called electric field space of charge or charged body. The field which is responsible for this force is called electric field.
- 4. 4 Electric Field Electric Field ~ D We generally define the electric field for vacuum or air medium and permit- tivity of air or vacuum is taken as ε0. Now assume that the electric field is measured in a medium of dielectric constant K and permittivity of ε(kε0), then the electric field in the medium is ~ E = 1 4πε Q r2 (1.2) Here the quantity 1 4π Q r2 is same for all mediums and the electric field varies by permittivity of the medium. Assuming this quantity as a new electric field vector, defined by ~ D, then the above equation becomes ~ E = ~ D ε (1.3) Or ~ D = ε ~ E (1.4) The dimensions of D are QL−2 , and the SI units are Cm−2 . This may seem to be rather trivial, but it does turn out to be more important than it may seem at the moment. Intensity of Electric Field Assume a unit positive experimental charge, q0, is placed near to another charge Q. Force experienced by experimental charge is known as electric field at that point due to charge Q. ~ Fk q0 = k Q r2 or ~ E = 1 4πε0 Q r2 (1.5) Direction of electric field is directed away from positive charge and directed to negative charge. Solved Problem 1.1 A charged particle having charge of 2.00 × 10−13 C is placed in an electric field of intensity 1.39 × 10−3 N/C. Find the force on it.
- 5. 1.1. ELECTRIC FIELD 5 Solution Force on charge placed in external electric field is given by q ~ E. Hence force on the charge 2.00 × 10−13 C in electric field 1.39 × 10−3 N/C is F = 2 × 10−13 × 1.39 × 10−3 It gives force on charge of magnitude 2.78 × 10−16 N. Electric Field Line Electric field lines originated from positive charge and ended at negative charge. These lines make open path. A unit positive experimental charge always follow path on electric field line when it is set free to move. Characteristics of Electric Field Line Electric field lines have following characteristics: - 1. Electric field lines generated from positive charge (source) and end at negative charge (sink). This statement can be redesigned as “the electric field lines emanate from higher potential source and end at the lower potential sink”. 2. Electric field lines make open curve. 3. Electric field lines never intersect to each other. 4. Electric field lines are appear to enter into a surface normally. 5. Electric field lines never enter into conductor’s surface. ~ E (F : 1) ~ E (F : 2) ~ E (F : 3) ~ E (F : 4) Figure 1.1: Interaction between electric field and conductor’s surfaces.
- 6. 6 Electric Field Electric field lines are originated from positive charge and ended at negative charge. 1.1.2 Electric Dipole It is a system of two unlike point charges of equal magnitude placed at finite distance. The line joining to the charges of an electric dipole is known as its axis and its distance is assumed as 2d or 2l, i.e. multiple of 2. +q −q 2l Figure 1.2: Electric dipole. Electric dipole moment ~ p is p = q2l where 2l is the displacement vector pointing from the negative charge to the positive charge. SI unit of electric dipole is C-m. The electric dipole moment vector ~ p points from the negative charge to the positive charge. An idealization of this two-charge system is the electrical point dipole consisting of two (infinite) charges only infinitesimally separated, but with a finite ~ p. 1.1.3 Electric Field Due To Electric Dipole There are two symmetrical position of point around the electric dipole where intensity of electric field is to be calculate. These are (i) axial position and (ii) equatorial position.
- 7. 1.1. ELECTRIC FIELD 7 Electric Field At Axis of Electric Dipole +q −q E+q E−q bc C 2l r Figure 1.3: Electric field at point due to electric dipole. The field around charges is Figure 1.4: Electric field lines at test point charge due to electric dipole. From figure (1.3) a point C is at distance r from the center of electric dipole in dipole axis. 2l is length of electric dipole. At point C electric field intensity due to positive charge of electric dipole is ~ E+q = 1 4πε0 | + q| (r + l)2 Away from electric dipole. (1.6) and due to negative charge of electric dipole is ~ E−q = 1 4πε0 | − q| (r − l)2 Towards electric dipole. (1.7) Direction of electric fields due to positive charge is away from center of electric dipole and due to negative charge is towards center of electric dipole. Both directions are opposite and parallel then resultant electric field due to electric dipole at point C is equal to | ~ E+q|−| ~ E−q|. Hence resultant electric field from
- 8. 8 Electric Field equations (1.6) and (1.7) at point C is ~ E = 1 4πε0 | + q| (r + l)2 − 1 4πε0 | − q| (r − l)2 = 1 4πε0 q (r + l)2 − q (r − l)2 = q 4πε0 (r − l)2 − (r + l)2 (r2 − l2)2 = q 4πε0 r2 + l2 − 2rl − r2 − l2 − 2rl (r2 − l2)2 Or ~ E = q 4πε0 −4rl (r2 − l2)2 (1.8) Equation (1.8) is applicable when 2l is not negligible with respect to distance r. If l r then l2 r2 so leaving negligible value l2 hence ~ E = q 4πε0 −4rl r4 or ~ E = 1 4πε0 −2q.2l r3 Using p = q.2l where p is electric dipole moment. ~ E = 1 4πε0 −2p r3 (1.9) This is equation for electric field at point C due to electric dipole. −p shows that electric field at point C due to negative charge is greater than electric field at point C due to positive charge.
- 9. 1.1. ELECTRIC FIELD 9 At Equatorial Position of Electric Dipole r E−q E+q 2l bc C −q P θ +q Q E+q E−q Eh +q + Eh −q Ev +q Ev −q bc C θ Figure 1.5: Electric field due to electric dipole in equatorial position. Figure 1.6: Electric field lines at test point charge due to electric dipole. From first part of figure (1.5), point C is at distance r from the center of the electric dipole. 2l is length of electric dipole. At point C electric field intensity due to positive charge of electric dipole is ~ E+q = 1 4πε0 | + q| (QC)2 (1.10) and due to negative charge of electric dipole is ~ E−q = 1 4πε0 | − q| (PC)2 (1.11) Direction of electric field due to positive charge is from Q to C and due to negative charge is from C to P. From second part of figure (1.5) ~ E1
- 10. 10 Electric Field has two parts, one ~ E1 cos θ parallel to horizontal axis towards Q to P and ~ E1 sin θ vertically upward. Similarly ~ E2 has two parts, one ~ E2 cos θ parallel to horizontal axis towards Q to P and ~ E2 sin θ vertically downward. Vertically downward and upward parts are parallel and opposite in direction hence they will cancel each other. So total Electric field at point C is ~ E = | ~ E1| cos θ + | ~ E2| cos θ. Substituting values of ~ E1 and ~ E2 the electric field at C will be ~ E = 1 4πε0 +q (QC)2 + 1 4πε0 +q (PC)2 cos θ Where from figure (??), QC = PC = p (r2 + l2) and cos θ = l √ (r2+l2) . Hence ~ E = 1 4πε0 +q (r2 + l2) + +q (r2 + l2) l p (r2 + l2) or ~ E = 1 4πε0 2q l (r2 + l2)3/2 (1.12) if l r then l2 r2 so leaving negligible value l2 hence ~ E = 1 4πε0 2q l (r2)3/2 Using p = q.2l ~ E = 1 4πε0 n p r3 o Where p is electric dipole moment. ~ E = 1 4πε0 p r3 (1.13) This is equation for electric field at point C due to electric dipole in equatorial axes. This is half of the electric field at axial point C given by equation (1.9). Solved Problem 1.2 Two charges 2µC and 3µC are placed at finite distance of 2cm. Find the electric field intensity at distance 10cm in equatorial position (perpendicular bisector of axis of dipole) of the charge system. Solution
- 11. 1.1. ELECTRIC FIELD 11 r 10cm E2 E3 2cm bc C 2µC P θ 3µC Q E2 E3 Ev 2 + Ev 2 EH 3 EH 2 bc C Electric field at the point C due to charges 2µC and 3µC are E2 = 1 4πε0 × 2 × 10−6 r2 and E3 = 1 4πε0 × 3 × 10−6 r2 Vector directions of these two electric field lines are away from the two charges along the line of axis joining charges to the point C. The magnitudes of electric field intensities E2 and E3 are unequal hence, these field lines should be resolved in horizontal and vertical components. E2h = 1 4πε0 × 2 × 10−6 r2 × cos θ E2v = 1 4πε0 × 2 × 10−6 r2 × sin θ and E3h = 1 4πε0 × 3 × 10−6 r2 × cos θ E3v = 1 4πε0 × 3 × 10−6 r2 × sin θ Horizontal components will be subtracted while vertical components will be added. Hence Eh = − 1 4πε0 × 1 × 10−6 r2 × cos θ Ev = 1 4πε0 × 5 × 10−6 r2 × sin θ
- 12. 12 Electric Field Now r, cos θ and sin θ are given by r = √ 101 × 10−2 m cos θ = 1 √ 101 sin θ = 10 √ 101 Substituting these values in horizontal and vertical components of electric fields Eh = − 9 × 107 101 √ 101 and Ev = 450 × 107 101 √ 101 Now net electric field is given by E = q E2 h + E2 v It gives E = 4.43 × 106 N/C bc The direction of this electric field is tan θ = Ev Eh It gives angle θ = 88.89◦ from negative x-axis.
- 13. 1.1. ELECTRIC FIELD 13 Solved Problem 1.3 An arc is substanding angle θ at its center. If charge density in the arc is λ then prove that magnitude of electric field at its center is E = 1 4πε0 2λ sin(θ/2) R Also find the direction of electric field in your answer. Solution θ E bc dl φ Eh A small element dl in the arc is producing electric field E at its center is given by E = 1 4πε0 λdl R2 Due to symmetry of arc, only horizontal component of electric field shall be effective, hence Eh = E cos φ Here φ is instantaneous angular direction of electric field from horizontal. Now Eh = 1 4πε0 λdl R2 cos φ Again for arc element dl = R dφ and Eh = 1 4πε0 λdφ R cos φ Integrating it for the angular limit of arc from −θ/2 to θ/2, electric field is Eh = 1 4πε0 2λ sin(θ/2) R Electric field will be in outward direction along the equatorial axis of the arc.
- 14. 14 Electric Field Geometrical Approach 1 Two like charges, Q1 and Q2 are placed at points (x1, y1) and (x2, y2) respectively. As both charges are like, hence neutral point will be in between these two coordinate points. Let neutral point is at (x, y). A, positive unit test charge is placed here to get the electric field intensity. If electric field intensity at this point is zero then horizontal and vertical components of electric field intensities due to both charges must be equal and in opposite direction as shown in the figure. x y b Q1 (x1, y1) b Q2 (x2, y2) b (x, y) ExQ1 EyQ1 ExQ2 EyQ2 Electric field at at neutral point due to charge Q1 is EQ1 = 1 4πε0 Q1 r2 1 (1.14) Here r1 = p (x1 − x)2 + (y1 − y)2. Gradient of these two points (x, y) and (x1, y1) is tan θ1 = y1 − y x1 − x Horizontal components of electric field due to charge Q1 is ExQ1 = 1 4πε0 Q1 r2 1 | cos θ1| (1.15) Here cos θ1 is cos θ1 = |x1 − x| p (x1 − x)2 + (y1 − y)2 Similarly vertical component of electric field due to charge Q1 is EyQ1 = 1 4πε0 Q1 r2 1 | sin θ1| (1.16) 1 Needs Approval
- 15. 1.1. ELECTRIC FIELD 15 Here sin θ1 is sin θ1 = |y1 − y| p (x1 − x)2 + (y1 − y)2 The horizontal and vertical component of electric field due to charge Q2 are ExQ2 = 1 4πε0 Q2 r2 2 | cos θ2| (1.17) and EyQ2 = 1 4πε0 Q2 r2 2 | sin θ2| (1.18) Where cos θ2 and sin θ2 are cos θ2 = |x − x2| p (x − x2)2 + (y − y2)2 and sin θ2 = |y − y2| p (x − x2)2 + (y − y2)2 At neutral point ExQ1 = ExQ2 and EyQ1 = EyQ2 Hence 1 4πε0 Q1 r2 1 | cos θ1| = 1 4πε0 Q2 r2 2 | cos θ2| Or Q1 Q2 = r2 1 r2 2 ×
- 19. cos θ2 cos θ1
- 23. (1.19) Similarly, equating vertical components, we have Q1 Q2 = r2 1 r2 2 ×
- 27. sin θ2 sin θ1
- 31. (1.20) Comparing above two relations, we get | tan θ1| = | tanθ2| (1.21)
- 32. 16 Electric Field This will gives relation between x and y, if two coordinates are known. Ac- cording to relation x1 − x y1 − y = x − x2 y − y2 (1.22) Now substitute value of x or y in any of two equations 1.14 or 1.20 we will get the value of x. Now we can calculate value of y from relation 1.22. 1.1.4 3-Dimensional Electric Field Problems y z x ~ r1 ~ E1 ~ r2 ~ E2 ~ E bc C bc A bc B ~ E1 ~ E2 ~ E bc Let two point charges ±Q1 and ±Q2 are placed at two points P and Q whose coordinates are (x1, y1, z1) and (x2, y2, z2) respectively. A positive test charge is placed at third point C of coordinate (x, y, z) where electric field is to be calculated. Now, electric field at C due to charge at point P is ~ E1 = 1 4πε0 ±Q1 |~ r1|3 ~ r1 (1.23)
- 33. 1.1. ELECTRIC FIELD 17 2 Similarly electric field at C due to charge at point Q is ~ E2 = 1 4πε0 ±Q2 |~ r2|3 ~ r2 (1.24) The values of ~ r1 and ~ r2 are ~ r1 = (x − x1)î + (y − y1)ĵ + (z − z1)k̂ (1.25) and ~ r2 = (x − x2)î + (y − y2)ĵ + (z − z2)k̂ (1.26) respectively. The magnitude of vectors ~ r1 and ~ r2 are r1 = p (x − x1)2 + (y − y1)2 + (z − z1)2 (1.27) and r2 = p (x − x2)2 + (y − y2)2 + (z − z2)2 (1.28) respectively. Net electric field at point C is ~ E = ~ E1 + ~ E2 (1.29) Substituting the values from equations 1.23 and 1.24 we have ~ E = 1 4πε0 ±Q1 |~ r1|3 ~ r1 + 1 4πε0 ±Q2 |~ r2|3 ~ r2 (1.30) 2 We know that the electric field E = 1 4πε0 Q r2 is a vector quantity and in vector form it is written as ~ E = 1 4πε0 ±Q r2 r̂ For a given vector, unit vector r̂ is given by r̂ = ~ r r So the vector form of electric field is ~ E = 1 4πε0 Q r3 ~ r
- 34. 18 Electric Field On simplification ~ E = 1 4πε0 ±Q1 |~ r1|3 ~ r1 + ±Q2 |~ r2|3 ~ r2 (1.31) This is the required value of electric field at the point C. Solved Problem 1.4 A student wants to use the properties of vectors to measure the electric field intensity at a point C. She takes two positive point charges of magnitudes 3µC and 4µC respectively and places them at points A and B respectively. She measures the magnitude of electric field at the point C. Find the magnitude of electric field at point C. Also find the direction of electric field. The point coordinates are A=(3, −1, 5), B=(3, 0, 3) and C=(1, 0, 3). Solution The vector ~ AC and ~ BC are ~ AC = (1 − 3)î + (0 − (−1))ĵ + (3 − 5)k̂ and ~ BC = (1 − 3)î + 0ĵ + (3 − 3)k̂ And magnitudes of these vectors are | ~ AC| = 3 and | ~ BC| = 2. Now electric field vector at point C is ~ E = 1 4πε0 3 × 10−6 | ~ AC|3 ~ AC + 4 × 10−6 | ~ BC|3 ~ BC # Substituting the values of vectors and their magnitudes ~ E = 10−6 4πε0 3 27 (−2î + ĵ − 2k̂) + 4 8 (−2î) On simplification ~ E = 10−6 4πε0 −11 9 î + 1 9 ĵ + −2 9 k̂ 1. Magnitude of the electric field at the point C in unit of N/C is obtained by finding the magnitude of above equation. Hence E = 42 3 × 103 2. The direction of electric field at the point is direction of unit vector of ~ E obtained by using relation ~ E/| ~ E|.
- 35. 1.1. ELECTRIC FIELD 19 Solved Problem 1.5 Three point charges 2µC, −2µC and −2µC are placed at coordinates (0, 0), (0, 2) and (3, 0) respectively. Find the electric field strength and its direction at coordinate point (4, 1). Solution 0 1 2 0 1 2 3 î ĵ A B C O r1 r2 r3 bc b b bc (a) 0 1 2 0 1 2 3 ~ Ex ~ Ey A B C O bc b b EA EB EC bc (b) ~ Ex ~ Ey EHA EHB EHC EV A EV B EV C bc (c) The vectors ~ AO, ~ BO and ~ CO are ~ AO = 4î + ĵ ~ BO = 4î − ĵ and ~ CO = î + ĵ And magnitudes of these vectors are | ~ AO| = √ 17, | ~ BO| = √ 17 and ~ CO| = √ 2. Now electric field vector at point O is ~ E = 2 × 10−6 4πε0 1 17 √ 17 (4î + ĵ) + −1 17 √ 17 (4î − ĵ) + −1 2 √ 2 (î + ĵ) 1. On simplification ~ E = 2 × 10−6 4πε0 −1 2 √ 2 î + 2 17 √ 17 − 1 2 √ 2 ĵ 2. The angular direction of electric field with respect to x-axis is tan θ = 2 17 √ 17 − 1 2 √ 2 −1 2 √ 2
- 36. 20 Electric Field Here, numerator and denominator are negative separately, hence the direc- tion of electric field will be towards third quadrant at point O. Now the angule value of direction of electric field from x-axis in thrid quadrant is θ = tan−1 1 − 4 √ 2 17 √ 17 # 1.2 Numerical Computation 1.2.1 Electric Field Electric field lines are continuous lines emerge from positive charge and end at negative charge. Electric field intensity at a point due to a point charge system is given by ~ E = 1 4πε0εr q ~ r |~ r|3 A two charge particles system is shown in given below. bc q1 b q2 b P(x, y) ~ r1, θ1 ~ r2, θ2 A virtual Gaussian surface of equal electric field intensity is drawn around this charge system. Every point on this Gaussian surface has equal strength of electric field. A point P(x, y) is chosen on this surface. In the given figure only a portion of the Gaussian surface is shown.
- 37. 1.2. NUMERICAL COMPUTATION 21 bc q1 b q2 b P(x, y) r1, θ1 ~ r2, θ2 Electric field on this surface is given by ~ E = ~ E1 + ~ E2 Flux through this surface is given by product of electric field an its area. Surface area of this portion of the Gaussian surface is not commutable as the surface may be symmetric or unsymmetrical. b q2 b P(x, y) ~ r2, θ2 S1 S2 ~ r ~ r2 − ~ r2 cos θ2 But for small surface area, we can consider it as spherical. The radius of the circular cross section of this surface is given by ~ r2 = (~ r2 + ~ r2 cos θ2)(~ r2 − ~ r2 cos θ2) On simplification ~ r2 = ~ r2 2 − ~ r2 2 cos2 θ2 It gives the area of the small surface area dS1 = π(~ r2 2 − ~ r2 2 cos2 θ2)
- 38. 22 Electric Field Now flux in this surface is dφ = E × dS1 Or dφ = 1 4πε0εr × q2 ~ r2 2 × π(~ r2 2 − ~ r2 2 cos2 θ2) It gives dφ = 1 4ε0εr q2(1 − cos2 θ2) If θ ≈ 0 then (1 − cos2 θ2) = 2(1 − cos θ2). Now dφ = 1 2ε0εr q2(1 − cos θ2) It is clear that flux over the given surface is depend only on the charge and angular position of point from the charge. Now we can take that q2(1 − cos θ2) = K Where K is a constant value. If there are n charge system enclosed by virtual Gaussian surface then i=n X i=1 qi(1 − cos θi) = K1 Or i=n X i=1 qi cos θi = K2 cos θ for a point charge at placed (x1, y1) with respect point (x, y) is given by cos θ = x − x1 p (x − x1)2 + (y − y1)2 Hence for n point charge system i=n X i=1 qi x − xi p (x − xi)2 + (y − yi)2 = K2 Now we can calculate all possible points for different K2 of electric field lines for which above relation is satisfied. Now the path of these points is the trace of electric field line. C program for generation of tracing points of electric potential surface of two charge system at [1C, −2x, 0y] and [−1C, 2x, 0y] is given below.
- 39. 1.2. NUMERICAL COMPUTATION 23 ✞ 1 #include stdio.h #include stdlib.h 3 #include math.h #define bufsize 100 5 char r[bufsize ]; FILE *fp; 7 int n; 9 int main(int argc , char** argv) { float x, y, k; 11 fp = fopen(a.txt, wb); for (x = -5; x 5; x = x + 0.1) { 13 for (y = -5; y 5; y = y + 0.1) { k = (x + 2) / pow((x + 2)*(x + 2) + y*y, 0.5) 15 - (x - 2) / pow((x - 2)*(x - 2) + y*y, 0.5); if(k 0.2409 k 0.25001) { 17 n = sprintf (r, %2.2ft%2.3ft%2.3fn, k, x, y); fwrite(r, sizeof (char), n, fp); 19 } } 21 } fclose(fp); 23 return (EXIT_SUCCESS); } ✌ ✆ A tabulted data table is saved in the file ‘a.txt’. Solved Problem 1.6 Find the tabulated data for electric field lines for charge system 1C and −0.5C placed at (−1, 0) and (2, 0) respectively. Solution ✞ #include stdio.h 2 #include stdlib.h #include math.h 4 #define bufsize 100 char r[bufsize ]; 6 FILE *fp;
- 40. 24 Electric Field int n; 8 int main(int argc , char** argv) { 10 float x, y, k; fp = fopen(a.txt, wb); 12 for (x = -12; x = 12; x = x + 0.01) { for (y = -12; y = 12; y = y + 0.01) { 14 k = (x + 1) / pow((x + 1)*(x + 2) + y*y, 0.5) - (x - 2) / pow((x - 2)*(x - 2) + y*y, 0.5); 16 if (k 0.295 k 0.30) { n = sprintf (r, %2.2ft%2.3ft%2.3fn, k, x, y); 18 fwrite(r, sizeof (char), n, fp); } 20 } } 22 fclose(fp); return (EXIT_SUCCESS); 24 } ✌ ✆ 1.2.2 Electric Potential Electric potential surfaces are spherical surface around charges having radii of potential magnitude at a point on the surface. Electric potential at a point due to a point charge is given by V = 1 4πε0εr q r A two charge particles system is shown in given below. bc q1 b q2 b P(x, y) r1 r2
- 41. 1.2. NUMERICAL COMPUTATION 25 At a point P(x, y) electric potential is given by V = V1 + V2 and from potential definitions: V = 1 4πε0εr q1 r1 + q2 r2 It is clear that potential at a point is depend only on the sum of ratios of charge to distance. q1 r1 + q2 r2 = K Where K is a constant value. If there are n charges in a system then i=n X i=1 qi ri = K1 Here ri = p (x − xi)2 + (y − yi)2 Hence for n point charge system i=n X i=1 qi p (x − xi)2 + (y − yi)2 = K1 Now we can calculate all possible points for different K1 of electric poten- tial surfaces for which above relation is satisfied. Now the path of these points is the trace of electric field line. C program for generation of tracing points of electric potential surface of two charge system at [1C, −2x, 0y] and [−1C, 2x, 0y] is given below. ✞ #include stdio.h 2 #include stdlib.h #include math.h 4 #define bufsize 100 char r[bufsize ]; 6 FILE *fp; int n; 8 int main(int argc , char** argv) { 10 float x, y, k; fp = fopen(a.txt, wb); 12 for (x = -5; x 5; x = x + 0.1) { for (y = -5; y 5; y = y + 0.1) {
- 42. 26 Electric Field 14 k = 1 / pow((x + 2)*(x + 2) + y*y, 0.5) - 1 / pow((x - 2)*(x - 2) + y*y, 0.5); 16 if (k -1 k 1) { n = sprintf (r, %2.2ft%2.3ft%2.3fn, k, x, y); 18 fwrite(r, sizeof (char), n, fp); } 20 } } 22 fclose(fp); return (EXIT_SUCCESS); 24 } ✌ ✆ A tabulted data table is saved in the file ‘a.txt’. 1.3 Displacement Field Consider an infinite parallel plate capacitor placed in space (or in a medium) with no free charges present except on the capacitor. In SI units, the charge density on the plates is equal to the value of the ~ D field between the plates. I A ~ D · d ~ A = Qfree (1.32) −q +q ~ D Figure 1.7: Electric field in vacuum. Outside the capacitor, the fields of the two plates cancel each other and | ~ E| = | ~ D| = 0.
- 43. 1.4. ELECTRIC FORCE 27 ~ D ~ E ~ D −q +q Figure 1.8: Electric field in presence of material. If the space between the capacitor plates is filled with a linear homo- geneous isotropic dielectric with permittivity ε, the total electric field ~ E between the plates will be smaller than ~ D by a factor of ~ ε, i.e. | ~ E| = Qfree εA (1.33) Hence, we can say that ~ D is inducing electric field to a material and ~ E is induced electric field in the material. 1.4 Electric Force If a point charge is placed in the electric field of other charged body then they mutually interact with each other. The strength or intensity of interaction between them is known as the electric force. Suppose an experimental charge of +qC is placed in the field intensity ~ E then electric force on the experimental charge is F = q ~ E (1.34) 1.4.1 Electric Dipole in Electric Field External electric field exerts electric force on electric dipole when it is placed in field. Positive charge on electric dipole experience force in the direction of electric field and negative charge experience force in the opposite direction of electric field. Both forces on positive and negative charge made force couple. These forces are parallel and opposite in direction hence they tend to align electric dipole parallel to electric field.
- 44. 28 Electric Field F−q F+q P Q ~ E θ 2l 2l cos θ 2l sin θ Figure 1.9: Electric dipole in homogeneous electric field. Electric field lines Figure 1.10: Effect of external electric field on electric dipole charges. Solved Problem 1.7 Two point charges 2µC and 2.5µC are placed by a distance of 1.2 × 10−1 .m Find the point where electric field intensity is zero. Solution 2µC 2.5µC E2 E2.5 bc C 0.12m x bc Both charges are positive, hence electric field will be zero inside the charges on the line of axis joining the charges. For this case assume the
- 45. 1.4. ELECTRIC FORCE 29 null point is at distance of x from 2µC. Now for the null points 2 × 10−6 x2 = 2.5 × 10−6 (0.12 − x)2 On solving it x = 0.0631m right from the 2µC charge. Solved Problem 1.8 Two point charges 1.2µC and −3.25µC are placed by a distance of 12cm. Find the point from positive charge where electric field intensity is zero. Solution 1.2µC −3.25µC E1.2 E−3.25 bc C 0.12m x bc Both charges are unlike charges, hence electric field will be zero outside the charges on the line of axis joining the charges. For this case assume the null point is at distance of x from 1.2µC. Now for the null points 1.2 × 10−6 x2 = 3.25 × 10−6 (0.12 + x)2 On solving it x = 0.18m left from the 1.2µC charge. Electric dipole Moment in Homogeneous Electric Field From figure (??) an electric dipole is placed in external electric field inclined at θ angle with electric field. Force on positive charge is F+q = q ~ E along the direction of electric field and force on negative charge is F−q = q ~ E
- 46. 30 Electric Field opposite to the direction of electric field. These forces make force couple hence force moment on electric dipole is τ = q ~ E 2l sin θ Where 2l sin θ is perpendicular distance between direction of two forces F+q and F−q. In terms of electric dipole moment p = q2l, τ = p ~ E sin θ To find the direction of electric dipole moment, we place electric dipole in an external electric field freely. Electric dipole aligns itself along the direction of electric field. Now the direction of electric field is parallel to the direction of electric dipole moment. Generally its direction is from more negative charge to less negative charge. Electric dipole Moment in Inhomogeneous Electric Field δx P Q ~ E ~ E + ~ δE Figure 1.11: Electric dipole in inhomogeneous electric field. Consider a simple dipole consisting two charges +q and −q separated by small distance δx. The force on the charge −q is q ~ E in opposite direction of electric field and force on the charge +q is q( ~ E + ~ δE) along the direction of electric field. If p is electric dipole moment then there is a net force to the right of magnitude q ~ δE. Or Force = p ~ dE dx (1.35) If potential drop between two point P and Q is V volt then electric field along the direction of x − axis is Ex = − ∂V ∂x (1.36)
- 47. 1.4. ELECTRIC FORCE 31 This is also known as relation between electric field and electric potential. Work Done When Electric Dipole Is In Electric Field With reference from figure (1.12), Force on electric dipole when it is placed in external electric field is τ = p ~ E sin α. Work required to rotate it a small angle dα is dW hence dW = p ~ E sin α dα on integrating above equation within limit α = θ1 to α = θ2 total work is W = −p ~ E [cos α]θ2 θ1 ~ E → ~ p bcb θ dθ Figure 1.12: Work done in rotating electric dipole in an electric field. Applying the limits W = −p ~ E [cos θ2 − cos θ1] W will be maximum when θ2 = 180◦ → cos θ = −1 and θ1 = 0◦ → cos θ = 1. Now work done is W = −p ~ E(−1 − 1) = 2p ~ E Solved Problem 1.9 An electric dipole with dipole moment 4 × 10−9 C-m is aligned at 30◦ with the direction of a uniform electric field of magnitude 5 × 104 N/C. Calculate the magnitude of the torque acting on the electric dipole. Solution Torque on a dipole is given by τ = pE sin θ
- 48. 32 Electric Field Placing the values of parameters τ = 4 × 10−9 × 5 × 104 × sin 30◦ It gives τ = 104 Nm. 1.4.2 Oil Drop Experiment The Oil drop experiment was carried out by Robert Millikan and Harvey Fletcher in 1909 to measure the elementary electric charge. Apparatus It consists two plates placed one over other. These plates are connected with regulated voltage source. Electrically charged oil drops are inserted between the plates and let them fall. Electric field between two plates is adjusted as the electrostatic force on the charged drop in vertically upward direction is balanced by the gravitational force on the drop in vertically downward direction. Originally the whole system was in air, but it can be placed in vacuum chamber. The motion of oil drop is observed by microscope. ~ E Fe Fg bb (a) ~ E Fe Fg Fd Fd bb (b) Principle This experiment uses the relation form vacuum chamber Fg = Fe (1.37) and for air chamber Fe = w + Fd (1.38) Where w is the apparent weight and Fd is drag force. Measurement In Original Experiment The drop is allowed to fall in the absence of an electric field. If v is the terminal velocity then Fd = 6πrηv (1.39)
- 49. 1.4. ELECTRIC FORCE 33 The oil drop is floating in air medium and assumed a perfect sphere hence its apparent weight is ~ w = 4π 3 r3 (ρ − ρair)~ g (1.40) At terminal velocity, drop is not accelerating, hence net force on it is zero and equation (1.39) and (1.40) are equal. This give the radius of the drop. r2 = 9ηv 2g(ρ − ρair) (1.41) Now electric field is applied and it is adjusted as it applied electrostatic force on the charged oil drop in vertically upward direction then Fe = q ~ E (1.42) If apparent weight is balanced by electrostatic force then Fe = w and from equations (1.39) and (1.42) q ~ E = 4π 3 r3 (ρ − ρair)~ g (1.43) But mass of oil drop can not be measured easily, hence electric field is ad- justed as the drop starts moving in upward direction and attained a constant terminal velocity vt, so Fe − w = Fd and q ~ E − 4π 3 r3 (ρ − ρair)~ g = 6πη(~ r · ~ vt) (1.44) If two plates are separated by ~ d then electric field and electric potential relation V = ~ E · ~ d (1.45) and relation (1.44) becomes q V ~ d − 4π 3 r3 (ρ − ρair)~ g = 6πη(~ r · ~ vt) (1.46) This relation can be used to find the charge of oil drop. Measurement In Vacuum Experiment In vacuum chamber oil drop ex- periment, there is no drag forces, buoyancy forces etc. Hence for the vacuum
- 50. 34 Electric Field chamber oil drop experiment electrostatic force is balanced by gravitational force. So Fe = Fg (1.47) And q V ~ d = mg (1.48) This relation can be used to find the charge on oil drop. 1.5 Charge Density Electric field due to any point is E ∝ q r2 . Electric charge distribution is of different type, Linear charge distribution, surface charge distribution and volume charge distribution. Linear Charge Distribution Charge Q is distributed over a length of ∆l then linear charge distribution is ρ = Q ∆l . Surface Charge Distribution Charge Q is distributed over a surface area of ∆s then surface charge distribution is ρ = Q ∆s . Volume Charge Distribution Charge Q is distributed over a volume of ∆V then volume charge distribution is ρ = Q ∆V . 1.6 Motion of Charged Particle in Electric Field Moving charge produces both magnetic filed as well as electric field. When a charge enters into the external electric field it interacts with it and produce mutual interacting electric force. The motion of charge particle may be in uniform electric field or in non-uniform electric field. 1.6.1 Motion of Charge in Uniform Electric Field A charge of mass m and having charge q0 enters into external electric field with parallel to it, at normal and at any angle θ.
- 51. 1.6. MOTION OF CHARGED PARTICLE IN ELECTRIC FIELD 35 Parallel to Electric Field When a charge enters parallel to the external electric field an electric force exerted on the charge. The value of electric force is q0E. If there are no other force acting on the charge then this force is responsible for acceleration of the charge particle. From Newton’s Law of motion F = q0 ~ E ma = q0 ~ E a = q0 ~ Em (1.49) ~ F ~ E Figure 1.13: Force on charged particle by external electric field. Charge is travelling parallely to external electric field. If ~ E is uniform (that is, constant in magnitude and direction), then the acceleration of charge particle is constant. The direction of acceleration of charge depends on its nature. If charge is positive then it will accelerate along the direction of external electric field. If charge is negative then it will accelerate in the opposite direction of the external electric field. Normal To Electric Field Let a charged particle of mass m and charge q0 is entering into a field at an angle 90◦ with constant velocity v î along x axis. The direction of electric field is along y axis and is Eĵ. The acceleration on the charge particle is along the direction of electric field.
- 52. 36 Electric Field ~ F vî Eĵ y x Figure 1.14: Force on charged particle by external electric field. Charge entering into electric field normally. Let the width of electric field is x in which the charge particle experience the force and it displaced by y along y axis. Initially the velocity of charged particle along x axis is ~ vî and acceleration is zero. After entering into the electric field, acceleration is a and is given by equation (1.49) aĵ = eq m ĵ This acceleration displaced charged particle by y throughout the flight, hence d2 y dt2 ĵ = eq m ĵ Here t is the time for which charge particle flights through the external electric field. On integrating above relation velocity of charge particle along y axis becomes dy dt = eq m t (1.50) Integrating again above relation (1.50) the vertical displacement y is y = eq m t2 2 (1.51) The horizontal displacement x is given by x = vî × t, where time t is given by relation (1.51) x = vî × t (1.52)
- 53. 1.7. RELATIVE PERMITTIVITY 37 1.7 Relative Permittivity The force between two charges, given by coulomb’s law Fv = 1 4πε0 q1q2 r2 is measured when the medium between these two charges is air or vacuum. If the air or vacuum is replaced by another medium, the force between these two charged reduces. This is because the electric field intensity between two charges reduced due to reverse polarity in the medium, which opposes external electric field between charges. If the magnitude of charges and distance between them is not change then coulomb’s force is Fm = 1 4πε q1q2 r2 Here ε is the permittivity of the medium. Permittivity of a medium is the property of the medium which determines limit of force between two charges placed in that medium. The ratio of Fv Fm is εr = 1 4πε0 q1q2 r2 1 4πε q1q2 r2 εr = ε ε0 Here εr is the relative permittivity of the medium with respect to air or vacuum. Thus Relative permittivity is the ratio of coulomb force be- tween two charges placed in vacuum to the coulomb force between these two charges, placed in a medium when their magnitude and distance between them is not changes. 1.8 Electric Field Due To Charge Distribu- tion The previous relation are obtained by assuming that both experimental charge and active charge are taken as point charges. But in practice ac- tual charge bodies are not in point shape but they are in linear, surface or volume shapes. Following sections are illustrating the methods of finding electric field due to different shapes of bodies.
- 54. 38 Electric Field 1.8.1 Charged Rod At Axial Position A rod of finite length L is charged by charge Q. Charge density over rod is λ = Q L . Electric field at center O due to charge dq = λdx of element dx at distance x from center O is d ~ E = 1 4πε0 dq x2 x y x x0 L dx dq = λ dx O Figure 1.15: Electric field due to charged rod. + Figure 1.16: Effect of electric field of charged rod on a unit test charges. To find electric field at O due to total charge on rod is obtained by integration above relation in limit from x0 to x0 + L. Hence ~ E = 1 4πε0 Z x0+L x0 λdx x2 ~ E = − 1 4πε0 λ x x0+L x0 ~ E = − λ 4πε0 1 x0 + L − 1 x0
- 55. 1.8. ELECTRIC FIELD DUE TO CHARGE DISTRIBUTION 39 ~ E = − λ 4πε0 x0 − x0 − L (x0 + L)(x0) ~ E = 1 4πε0 λL x2 0 + Lx0 Putting values of λ above equation becomes ~ E = 1 4πε0 Q x2 0 + Lx0 (1.53) If x0 L then equation (1.53) becomes ~ E = 1 4πε0 Q x2 0 (1.54) Equation (1.54) is the desired relation for the electric field at any point on the axial line of the rod. At this position, whole rod behaves like a point charge Q. 1.8.2 Charged Rod On Equatorial Position x y x dx L r y ~ E bbc P θ O x y ~ E ~ Ev ~ Eh bbc Figure 1.17: Electric field due to charged rod in equatorial position.
- 56. 40 Electric Field + Figure 1.18: Effect of electric field of charged rod on a unit test charges in equatorial axis. From first part of figure (1.17) a point P is at distance y from the center of the rod. L is length of rod. Electric field intensity at point P due to electric charge on small element dx at distance y from point P is d ~ E = 1 4πε0 dq r2 Putting value λ = Q L equation becomes d ~ E = 1 4πε0 λdx x2 + y2 From second part of figure (1.17) there are two elements of electric field ~ E on is ~ E cos θ perpendicular upward and ~ E sin θ towards parallel to axis of rod. Horizontal components of electric field would cancelled each other hence electric field due to charged element dx is ~ dEy = ~ E cos θ so ~ dEy = 1 4πε0 λ dx x2 + y2 cos θ (1.55) Integrating equation (1.55) within limit from −L/2 to L/2 equation becomes ~ Ey = 1 4πε0 Z L/2 −L/2 λ dx x2 + y2 cos θ Again substituting value of cos θ above equation becomes ~ Ey = 1 4πε0 Z L/2 −L/2 λy dx (x2 + y2)( p x2 + y2)
- 57. 1.8. ELECTRIC FIELD DUE TO CHARGE DISTRIBUTION 41 On simplification it gives ~ Ey = 1 4πε0 Z L/2 −L/2 λy dx (x2 + y2)3/2 Substituting x = y tan θ and dx = y sec2 θ dθ and ignoring the limits upto final integration in above equation which gives ~ Ey = 1 4πε0 Z λy2 sec2 θ y3(tan2 θ + 1)3/2 dθ = λy 4πε0 Z sec2 θ y2(sec2 θ)3/2 dθ = λy 4πε0 Z sec2 θ y2(sec3 θ) dθ Or ~ Ey = λy 4πε0 Z 1 y2(sec θ) dθ = λ 4πε0 Z cos θ y dθ = λ 4πε0 1 y [sin θ] Substituting the value of sin θ Ey = λ 4πε0y x p x2 + y2 # Applying limits for x Ey = λ 4πε0y x p x2 + y2 #L/2 −L/2 On simplification it gives Ey = λ 4πε0y L/2 p (L/2)2 + y2 − −L/2 p (−L/2)2 + y2 #
- 58. 42 Electric Field Or ~ Ey = λ 4πε0y L p (L/2)2 + y2 # (1.56) If y L then equation (1.56) will be approximated to ~ Ey = λ 4πε0y L y or ~ Ey = 1 4πε0 Q y2 (1.57) Here Q = λL i.e. total charge on the rod. If L y then equation (1.57) becomes ~ Ey = 2λ 4πε0y (1.58) Equation (1.58) is similar to the electric field at axis of charged cylinder. 1.8.3 Charged Rod On Un-Symmetrical Position Electric field ~ E here is due to long charged wired of length L. Here symmetry is not existed. Small element dx of wire carrying charge q = Lλ where λ is charge density per unit length of wire. A point P is at distance R from element dx. r is perpendicular distance of point P from wire. Electric field dE is dE = 1 4πε0 λ dx R2 . x y R r ~ E x L A B D C bbc P θ1 θ θ2 x y ~ E Ex Ey bbc P Figure 1.19:
- 59. 1.8. ELECTRIC FIELD DUE TO CHARGE DISTRIBUTION 43 + Figure 1.20: Effect of electric field of charged rod on a unit test charges in non axial position. There are two components of this electric field which are dEx in horizontal direction and dEy is vertical direction. If dE makes an angle θ with wire then dEx = 1 4πε0 λ dx cos θ R2 (1.59) dEx = 1 4πε0 λ dx sin θ R2 (1.60) Here θ2, θ1 and r are constants then relation between r and x can be given by using triangle △APB we have AC + CB = L − x + r cot θ2 Here AC + CB = AB = r cot θ so L − x + cos θ2 = r cot θ Differentiating above equation we get dx = r csc2 θ dθ Now horizontal component from equation (1.59) dEx = 1 4πε0 λr cos θ csc2 θ dθ R2 From figure (1.19), r/R = sin θ which gives horizontal component of electric field ~ dE dEx = 1 4πε0 λ r cos θ dθ
- 60. 44 Electric Field Similarly vertical component is dEy = 1 4πε0 λ r sin θ dθ Values of dEx and dEy are integrated result of above two equations so Ex = Z θ2 θ1 dEx = Z θ2 θ1 1 4πε0 λ r cos θ dθ And Ex = 1 4πε0 λ r (sin θ2 − sin θ1) Similarly Ey = − 1 4πε0 λ r (cos θ2 − cos θ1) As Ex and Ey are normal to each other then resultant of these two com- ponents is square root of sum of squares of components. E = p E2 x + E2 y gives E = 1 4πε0 λ r [(sin θ2 − sin θ1)2 + (cos θ2 − cos θ1)2 ]1/2 (1.61) And tan θ = cos θ2 − cos θ1 sin θ2 − sin θ1 (1.62) This is the required solution. There are several conditions. (1) when θ2 = 90◦ then E = 1 4πε0 λ r p (2)(1 − sin θ1)1/2 (1.63) And tan θ = cos θ1 1 − sin θ1 (1.64) (2) when wire is too long then θ1 = 0 then E = 1 4πε0 λ r √ 2 (1.65) And tan θ = 45◦ (1.66)
- 61. 1.8. ELECTRIC FIELD DUE TO CHARGE DISTRIBUTION 45 1.8.4 Charged Ring At Position On Its Axis From first part of figure (1.21) a point P is at distance r from the center of the ring. Electric field intensity at point P due to electric charge on small element dl which is at distance r from point P is ~ dE = 1 4πε0 dq r2 Putting value dq = λR dφ where λ is charge density on ring and r = p R2 + y2 equation becomes ~ dE = 1 4πε0 λR dφ R2 + y2 Vertical component of above equation is ~ dEy = ~ E cos θ so ~ dEy = 1 4πε0 λR dφ R2 + y2 cos θ (1.67) Integrating equation (1.67) within limit from φ = 0 to φ = 2π equation becomes ~ Ey = 1 4πε0 Z 2π 0 λR dφ R2 + y2 cos θ dl dq R r y ~ E P bbc θ φ dφ O ~ E ~ Ex ~ Ey bbc x y Figure 1.21: Electric field due to charged ring in plain perpendicular axis: non-symmetrical and symmetrical view.
- 62. 46 Electric Field R y b bc P r ~ E θ ~ E sin θ ~ E cos θ ~ E Figure 1.22: Three dimensional projection of electric field. Again substituting value of cos θ above equation becomes ~ Ey = 1 4πε0 Z 2π 0 λR dφ y (R2 + y2)( p R2 = y2) ~ Ey = λRy 4πε0 1 (R2 + y2)3/2 Z 2π 0 dφ or ~ Ey = λRy 4πε0 1 (R2 + y2)3/2 2π Here Q = 2πRλ is total charge on ring. So ~ Ey = 1 4πε0 qy (R2 + y2)3/2 (1.68) Equation (1.68) is desired equation. 1.8.5 Charged Disk On Its Axis Assume a disk from figure (1.23), is formed of several rings of varying radius. Let A ring of radius x and width dx which has charge density σ, then charge on this imaginary ring is dq = (2πxdx)σ. A point P is at distance y from the center of the ring. Electric field intensity at point P due to electric charge on ring of radius x which is at distance r from point P is ~ dEy = 1 4πε0 dq y (x2 + y2)3/2
- 63. 1.8. ELECTRIC FIELD DUE TO CHARGE DISTRIBUTION 47 Putting value dq and integrating above equation within limit from r = 0 to r = R ~ Ey = 2πσy 4πε0 Z R 0 x dx x2 + y2 R r y ~ E P bbc θ x dx dq O ~ E ~ Ex ~ Ey bbc x y Figure 1.23: Electric field due to charged disk in plain perpendicular axis: non-symmetrical symmetrical view. Putting x2 + y2 = u and 2x dx = du in above equation ~ Ey = 2πσy 4πε0 Z R2+y2 y2 du/2 u3/2 Or ~ Ey = πσy 4πε0 u−1/2 −1/2 R2+y2 y2 Or ~ Ey = 2πσy 4πε0 y |y| − y R2 + y2 If y 0 then above equation becomes ~ Ey = 2πσ 4πε0 1 − y R2 + y2 (1.69) Similarly if y 0 then result is ~ Ey = 2πσ 4πε0 −1 − y R2 + y2 (1.70)
- 64. 48 Electric Field R y b bc P r ~ E θ ~ E sin θ ~ E cos θ ~ E Figure 1.24: Three dimensional projection of electric field. If y R then 1 − y R2 + y2 ≈ 1 2 R2 y2 Hence equation (1.69) gives ~ Ey = 2πσ 4πε0 1 2 R2 y2 (1.71) Putting Q = πR2 σ then equation (1.71) gives ~ Ey = 1 4πε0 Q y2 (1.72) At this condition disk behaves like a charge point. Electronic Theory of Sparking When an electric wire become loose in its terminal, it starts sparking. Sparking can be explained by electronic, electric field theory. Sparking starts in those connections or terminals where wire become loose and a non- conducting gap came to existence. Assume a two phase wire connection, in which one end of wire is connected with positive phase and second end of the wire is near to negative phase with small gap. Wire’s second end is positive with respect to negative phase and there exists an electric filed between wire and negative phase. Both connecting wire and phase wire have rough surfaces, in which there are several spikes with sharp nodes. We know that the electric field between two oppositely charge conductors attract to charge
- 65. 1.8. ELECTRIC FIELD DUE TO CHARGE DISTRIBUTION 49 carriers. Under this electrostatic attraction, electrons from negative phase are accumulates in the nodes of spikes. If potential difference between two phases is very large then electric field intensity between wire and negative phase is very strong and the accumulation of electrons at spike nodes is very very large. As the node’s surface is very small, charge density at node becomes so huge that these electrons start leaving the node’s surface and reaches to the wire’s end. The electrons diffused with positive charge at wire’s end and a glow and energy generated. The charge dissipation between negative phase and wire’s end is very large and the medium air heated adiabatically. This expansion produces sound like thunder stroke of lightening. The combined phenomenon of glow, energy and sound is generally called sparking. 1.8.6 Cavity in a Non-Conducting Sphere Let a sphere of radius a b + z has a cavity of radius b at distance of z from the center of the sphere. Electric field at the cavity is the vector subtraction of the electric field of the sphere at the center of cavity and electric field of the sphere of size of the cavity. The charge density on the sphere is ρ. b b a z b b b z r′ r bc P î ĵ Vector of the point at any point inside the cavity is ~ r = xî + yĵ + zk̂ = rr̂ Using Gausss Law to calculate the electric field at the point at distance ~ r is
- 66. 50 Electric Field b r bc P I ~ E · d ~ A = 1 ε0 Qenc Or I ~ E · d ~ A = 1 ε0 4 3 πr3 ρ Or E 4πr2 = 1 ε0 4 3 πr3 ρ Or E = ρ 3ε0 r In vector form, electric field is ~ E = ρ 3ε0 rr̂ (1.73) The coordinate of the point P for cavity is ~ r′ = x′ î′ + y′ ĵ′ + z′ k̂′ = r′ r̂′ Applying the Gauss law for this cavity assuming that the charge density is −ρ as charge is removed from the cavity. b b r′ bc P
- 67. 1.9. ELECTRIC FORCE ON SURFACE OF CONDUCTOR 51 ~ E′ = −ρ 3ε0 r′ r̂′ (1.74) Superposition of both the electric fields ~ Ef = ~ E + ~ E′ On simplification ~ Ef = ρ 3ε0 rr̂ − ρ 3ε0 r′ r̂′ z r′ r î ĵ From the vector rule zĵ + r′ r̂′ = rr̂ Substituting it for final electric field at the cavity gives the final result as ~ Ef = ρ 3ε0 (zĵ + r′ r̂′ ) − ρ 3ε0 r′ r̂′ Or ~ Ef = ρ 3ε0 zĵ (1.75) This is the required derivative. 1.9 Electric Force on Surface of Conductor P Figure 1.25: Cross-section view of charged sphere.
- 68. 52 Electric Field A sphere of large radius R has charge q, homogeneously divided all over its surface. Surface of sphere of large radius is acts like a thick sheet. Electric field just outside the unit elemental surface of sphere is given by E = σ 2ǫ0 Force on the charge on unit surface area is F = σ × σ 2ǫ0 Electrostatic pressure on the unit surface of large sphere is P = F/A and P = × σ2 2ǫ0 From the definition of electric field E = σ ε0 From this relation, pressure becomes P = ε0 E2 2 (1.76) Solved Problem 1.10 A bubble soap is charged by charge q. Find the pressure inside the bubble. Solution Ps Ps (a) Ps Ps Pe (b) Figure 1.26: Cross-section view of charged bubble soap. (a) Surface pressure in both surfaces of bubble. (b) Electrically charged bubble and electrostatic pressure along-with surface pressure.
- 69. 1.9. ELECTRIC FORCE ON SURFACE OF CONDUCTOR 53 In a charged bubble, electrostatic pressure is outward while hydrostatic tension pressure is inward. Net pressure on the bubble is algebraic sum of these pressures. Pressure due to surface tension in soap bubble is 4T/r while electrostatic pressure is ε0E2 /2. r is radius of the bubble. Now Pint = 4T r − q2 32π2ε0r4 (1.77)