![dE =
σ ୢୱ
ସగఌమ
cosߠ az (1.28)
ds = π[(x + dx)2
– x2
]
ds = π[x2
+ dx2
+ 2xdx – x2
] = 2π x dx (neglecting dx2
term)
Substituting ds in eq. (1.28) , we get
dE =
σ ଶπ୶ୢ୶
ସగఌమ cosߠ az
dE =
σ ୶ୢ୶
ଶఌమ
cosߠ az (1.29)
From above figure we can write
Tanߠ = x/h
x = h tanߠ
(1.30)
dx = h sec2
ߠ dߠ
(1.31)
cosߠ = h/r
r = h/ cosߠ = h secߠ
(1.32)
Substituting eqs. (1.30), (1.31) and (1.32) in (1.29) we get,
dE =
σ ( ୦ ୲ୟ୬θ)(୦ ୱୣୡమθ ୢθ)
ଶఌ୦ ୱୣୡθమ cosߠ az
dE =
σ
ଶఌ
sinߠ dߠ az
On integrating
ࡱ =
ߪ
2ߝ
න ߠ݊݅ݏ
ఈ
݀ߠ ࢇࢠ
ࡱ =
ߪ
2ߝ
ሾ−ܿߠݏሿ
ఈ
ࢇࢠ
ࡱ =
ߪ
2ߝ
(1 − ܿࢇ)ߙݏࢠ (1.33)
From figure 1.7
ܿߙݏ =
ℎ
ඥ(ܽଶ + ℎଶ)](https://image.slidesharecdn.com/emfuniti-201114062820/85/Emf-unit-i-12-320.jpg)
Emf unit i
- 1. UNIT - I ELECTROSTATICS Introduction Charges at rest produce Static Electric Field or Electrostatic field. Field: It is the existing space in particular area due to some elements. 1.1) Coulomb’s Law Coulomb stated that the force between two point charges separated in a vacuum or free space by a distance which is large compared to their size is (i) proportional to the charge on each (ii) inversely proportional to the square of the distance between them (iii) directed along the line joining the charges ߙ ܨ ܳଵܳଶ ܴଶ ଵଶ Fig. 1.1 If Q1 and Q2 have like signs the vector force ࡲ on Q2 is in the same direction as ࡾ ܨ = ݇ ܳଵܳଶ ܴଶ ଵଶ (1.1) Where Q1 and Q2 are the positive or negative quantities of charge, R12 is the separation, and k is proportionality constant. If the International System of Units (SI) is used. In SI units, charges Q1 and Q2 are in coulombs (C), the distance R12 is in meters (m), and the force F is in newtons (N) so that k = 1/4πεo. The constant εo is known as the permittivity of free space (in farads per meter) and has the value εo= 8.854×10-12 ≈ ଵషవ ଷగ F/m k = 1/(4πεo) = 9 × 109 m/F Thus Eq. (1.1) becomes
- 2. ܨ = 1 4ߨߝ ܳଵܳଶ ܴଶ ଵଶ (1.2) If point charges Q1 and Q2 are located at points having position vectors ࢘ and ࢘, then the force F2 on Q1 due to Q2, shown in Figure 1.1, is given by ࡲ = 1 4ߨߝ ܳଵܳଶ ܴଶ ଵଶ ࢇࡾ (1.3) Where ࢇࡾ = ࡾ |ࡾ| (1.4) ࡾ = ࢘ − ࢘ ܴଵଶ = |ࡾ| By substituting eq. (1.4) into eq. (1.3), we may write eq. (1.3) as ࡲ = 1 4ߨߝ ܳଵܳଶ ܴଷ ଵଶ ࡾ (1.5) It is worthwhile to note that The force ࡲ on Q1 due to Q2 is given by ࡲ = |ࡲ|ࢇࡾ = |ࡲ|(−ࢇࡾ ) = −ࡲ ࡲ = −ࡲ (1.6) Problems 1. The charge Q2 = 10 µc is located at (3,1,0) and Q1 = 50 µc is located at (-1,1,-3). Find the force on Q1. Ans. R21 = (-1-3) i + (1-1) j + (-3-0) k = -4i-3k | R21| = ඥ(−4)ଶ + (−3)ଶ = 5 ࡲ = 1 4ߨߝ ܳଵܳଶ ܴଷ ଶଵ ࡾ ࡲ = 9 × 10ଽ 10 × 10ି × 50 × 10ି 5ଷ (−4i − 3k) ࡲ = −. − . ૡ 2. A point charge Q1 = 300 µc located at (1,-1,-3) experiences a force F1= 8i-8j+4k due to a point charge Q2 at (3,-3,-2)m. Determine Q2. Ans. R21 = (1-3) i + (3-1) j + (2-3) k = -2 i +2 j - k
- 3. | R21| = ඥ(−2)ଶ + (2)ଶ + (−1)ଶ = 3 ࡲ = 1 4ߨߝ ܳଵܳଶ ܴଷ ଶଵ ࡾ (8i − 8j + 4k ) = 9 × 10ଽ ܳଶ × 300 × 10ି 3ଷ (−2 i + 2 j − k) Q2 = -40 µc. 1.2) Force due to N no. of charges: If we have more than two point charges, we can use the principle of superposition to determine the force on a particular charge. The principle states that if there are N charges Q1, Q2, Q3..................... QN located, respectively, at points with position vectors ࢘, ࢘, ࢘ … … … … … ࢘ࡺ , the resultant force ࡲ on a charge Q located at point(p) ࢘ is the vector sum of the forces exerted on Q by each of the charges Q1, Q2, Q3..................... QN. Hence: ࡲ = ࡲ + ࡲ + ⋯ + ࡲࡺ ࡲ = 1 4ߨߝ ܳܳଶ ܴଷ ଵ ࡾ + 1 4ߨߝ ܳܳଶ ܴଷ ଶ ࡾ + ⋯ + 1 4ߨߝ ܳܳଶ ܴଷ ே ࡾࡺ (1.7) ࡲ = ܳ 4ߨߝ ܳ ܴଷ ࡾ ே ୀଵ (1.8) ࡲ = 9 × 10ଽ ܳ ܳ ܴଷ ࡾ ே ୀଵ (1.9) Problems 1. Find the force on a 100 µc charge at (0,0,3)m. If four like charges of 20 µc are located on x and y axes at ±4m. Ans. R1p = (0-4) i + (0-0) j + (3-0) k = -4 i +3 k | R1p| = ඥ(−4)ଶ + (3)ଶ = 5 R2p = (0-0) i + (0-4) j + (3-0) k = -4 j +3 k | R2p| = ඥ(−4)ଶ + (3)ଶ = 5 R3p = (0+4) i + (0-0) j + (3-0) k = 4 i +3 k
- 4. | R3p| = ඥ(4)ଶ + (3)ଶ = 5 R4p = (0-0) i + (0+4) j + (3-0) k = 4 i +3 k | R4p| = ඥ(4)ଶ + (3)ଶ = 5 ࡲ = 9 × 10ଽ ܳ ܳ ܴଷ ࡾ ே ୀଵ ࡲ = 9 × 10ଽ 100 × 10ି × 20 × 10ି 5ଷ (−4 i + 3 k − 4 i + 3 k + 4 i + 3 k + 4 i + 3 k) ࡲ = . ૠ ࡷ ܰ 2. Two small diameter 10 gm dielectric balls can slide freely on a vertical plastic channel. Each ball carries a negative charge of 1 nc. Find the separation between the balls if the upper ball is restrained from moving. Ans. Gravitational force and coulomb force must be equal F2 = mg = 10 × 10-3 × 9.8 = 9.8× 10-2 N. ܨଵ = 1 4ߨߝ ܳܳଶ ܴଶ ଵ Therefore F1 = F2 9.8× 10-2 = ଽ×ଵవ (ଵ× ଵషల)మ ோమ R=0.302 m 1.3) Electric Field Intensity or Electric Field Strength (ࡱ): It is the force per unit charge when placed in electric field. Fig 1.2 The lines of force due to a pair of charges, one positive and the other negative
- 5. Fig 1.3 The lines of force due to a pair of positive charges An electric field is said to exist if a test charge kept in the space surrounding another charge, then it will experience a force. Thus, ࡱ = lim ொ→ ࡲ࢚ ܳ௧ (1.10) or simply ࡱ = ࡲ࢚ ܳ௧ (1.11) The electric field intensity ࡱ is obviously in the direction of the force ࡲ and is measured in newton/coulomb or volts/meter. The electric field intensity at point ࢚࢘ due to a point charge located at ࢘ is readily obtained from eq. (1.3) as ࡲ࢚ = 1 4ߨߝ ܳଵܳ௧ ܴଶ ଵ௧ ࢇࡾ࢚ ࡲ࢚ ܳ௧ = 1 4ߨߝ ܳଵ ܴଶ ଵ௧ ࢇࡾ࢚ ࡱ = 1 4ߨߝ ܳଵ ܴଶ ଵ௧ ࢇࡾ࢚ (1.12) 1.3.1) Electric field due to N no. of charges: If we have more than two point charges, we can use the principle of superposition to determine the force on a particular charge. The resultant force ࡲ on a charge Q located at point(p) ࢘ is the vector sum of the forces exerted on Q by each of the charges Q1, Q2, Q3..................... QN. Hence: ࡲ = ࡲ + ࡲ + ⋯ + ࡲࡺ From eq. (1.9)
- 6. ࡲ = 9 × 10ଽ ܳ ܳ ܴଷ ࡾ ே ୀଵ We know that ࡱ = ࡲ ܳ ࡱ = 9 × 10ଽ ܳ ܴଷ ࡾ ே ୀଵ (1.13) 1.3.2) Electric Fields Due to Continuous Charge Distributions: So far we have only considered forces and electric fields due to point charges, which are essentially charges occupying very small physical space. It is also possible to have continuous charge distribution along a line, on a surface, or in a volume as illustrated in figure 1.2. Fig. 1.4 volume charge distribution and charge elements It is customary to denote the line charge density, surface charge density, and volume charge density by λ (in C/m), σ (in C/m2 ), and ρv (in C/m3 ), respectively. Line charge density: when the charge is distributed over linear element, then the line charge density is the charge per unit length. λ = lim ௗ→ ݀ݍ ݈݀ where dq is the charge on a linear element dl. Surface charge density: when the charge is distributed over surface, then the surface charge density is the charge per unit area.
- 7. σ = lim ௗ௦→ ݀ݍ ݀ݏ where dq is the charge on a surface element ds. Volume charge density: when the charge is confined within a volume, then the volume charge density is the charge per unit volume. ρ୴ = lim ௗ௩→ ݀ݍ ݀ݒ where dq is the charge contained in a volume element dv. 1.3.3) Electric field due to line charge: Consider a uniformly charged wire of length L m, the charge being assumed to be uniformly distributed at the rate of λ (linear charge density) c/m. Let P be any point at which electric field intensity has to be determined. Consider a small elemental length dx at a distance x meters from the left end of the wire, the corresponding charge element is λ dx. Divide the wire into a large number of such small elements, each element will render its contribution towards the production of field at P. Fig 1.5 Evaluation E due to line charge Let dE be field due to the charge element λ dx. It has a component dEx along x-axis and dEy along y-axis. dE = dEx ax + dEy ay
- 8. we know that ݀ܧ = λ dx 4ߨߝݎଶ (1.14) From figure 1.5 we can write, dEx = dE cosߠ (1.15) dEy = dE sinߠ (1.16) Therefore we can write, ݀ܧ௫ = λ cosθ dx 4ߨߝݎଶ (1.17) ݀ܧ௬ = λ sinθ dx 4ߨߝݎଶ (1.18) Substituting eqs. (1.17) and (1.18) in dE we get, ݀ࡱ = λ cosθ dx 4ߨߝݎଶ ࢇ࢞ + λ sinθ dx 4ߨߝݎଶ ࢇ࢟ (1.19) From figure 1.5 we write L1-x = h cotߠ (1.20) -dx = -h cosec2 ߠ dߠ (1.21) r = h cosecߠ (1.22) Substituting equations (1.20), (1.21) and (1.22) in 1.19, we get ݀ࡱ = λ cosθ dθ 4ߨߝℎ ࢇ࢞ + λ sinθ dθ 4ߨߝℎ ࢇ࢟ (1.23) The electric field intensity E due to whole length of the wire ࡱ = න ݀ࡱ ݀ߠ ࣂୀ࣊ିࢻ ࣂୀࢻ ࡱ = න λ cosθ dθ 4ߨߝℎ ࢇ࢞ + λ sinθ dθ 4ߨߝℎ ࢇ࢟ ൨ ݀ߠ ࣂୀ࣊ିࢻ ࣂୀࢻ
- 9. ࡱ = ߣ 4ߨߝℎ ൣࢇ ߠ݊݅ݏ࢞ − ܿࢇ ߠݏ࢟൧ ࢻ ࣊ିࢻ ࡱ = ߣ 4ߨߝℎ ൣ(ߙ݊݅ݏଶ − ߙ݊݅ݏଵ)ࢇ࢞ + (ܿߙݏଶ + ܿߙݏଵ) ࢇ࢟൧ ܰ ܥ (1.24) Case (i) If P is the midpoint, α1= α2 = α ࡱ = ߣ 4ߨߝℎ ൣ(ߙ ݊݅ݏ − ࢇ)ߙ ݊݅ݏ࢞ + (ܿߙ ݏ + ܿࢇ )ߙ ݏ࢟൧ ܰ ܥ ࡱ = ߣ 4ߨߝℎ ൣ2 ܿࢇ ߙ ݏ࢟൧ ܰ ܥ ࡱ = ߣ 2ߨߝℎ ܿࢇ ߙ ݏ࢟ ܰ ܥ (1.25) The direction of electric field intensity is normal to the line charge. The electric field is not normal to the line charge if the point is not at midpoint. Case (ii) As length tends to α1 = 0 and α2 = 0 from equation (1.24), we get ࡱ = ߣ 2ߨߝℎ ࢇ࢟ ܰ ܥ (1.26) 1.3.4) Electrical field due to charged ring: A circular ring of radius a carries a uniform charge λ C/m and is placed on the xy-plane with axis the same as the z-axis as shown in figure. Let dE be the electric field intensity due to a charge dQ. The ring is assumed to be formed by several point charges. When these vectors are resolved, radial components get cancelled and normal components get added. Therefore the direction of electric field intensity is normal to the plane of the ring. The sum of normal components can be written as ∞
- 10. Fig. 1.6 charged ring ࡱ = න ݀ࢇ ߙݏܿ ܧࢠ ࡱ = න ݀ܳ 4ߨߝܴଶ ܿࢇ ߙ ݏࢠ ࡱ = න ߣ ݈݀ 4ߨߝܴଶ ℎ ܴ ࢇࢠ ࡱ = ߣ ℎ 4ߨߝܴଷ ࢇࢠ න ݈݀ ࡱ = ߣ ℎ 4ߨߝܴଷ × 2ߨܽ ࢇࢠ ࡱ = ߣ ℎ 4ߨߝ√ܽଶ + ℎଶ ଷ × 2ߨܽ ࢇࢠ ࡱ = ߣ ܽ ℎ 2ߝ(ܽଶ + ℎଶ)ଷ/ଶ ࢇࢠ N C (1.27)
- 11. 1.3.5) Electrical field due to a charged disc: A disc of radius ‘a’ meters is uniformly charged with a charged density σ c/m2 . It is required to determine the electric field at ‘P’ which is at a distance h meters from the centre of the disc as shown in figure 1.7 Fig. 1.7 charged disc The disc is assumed to be formed by several rings of increasing radius. Consider a ring of radius x meters. Each ring is assumed to be formed by number of point charges. Let dE1 be the electric field intensity due to a charge dQ1 and dE2 is the electric field intensity due to a charge dQ2. Electric field due to one ring be obtained by adding normal components of dE1, dE2 ....... dEn. Therefore dE = (dE1 cosߠ + dE2 cosߠ + ....... + dEn cosߠ ) az dE = (dE1 + dE2 + ....... + dEn) cosߠ az dE = ( ௗொభ ସగఌమ + ௗொమ ସగఌమ + ....... + ௗொ ସగఌమ) cosߠ az dE = ௗொభାௗொమା⋯ାௗொ ସగఌమ cosߠ az dE = ௗொ ସగఌమ cosߠ az The total charge of the ring is σ ds which is equal to dQ
- 12. dE = σ ୢୱ ସగఌమ cosߠ az (1.28) ds = π[(x + dx)2 – x2 ] ds = π[x2 + dx2 + 2xdx – x2 ] = 2π x dx (neglecting dx2 term) Substituting ds in eq. (1.28) , we get dE = σ ଶπ୶ୢ୶ ସగఌమ cosߠ az dE = σ ୶ୢ୶ ଶఌమ cosߠ az (1.29) From above figure we can write Tanߠ = x/h x = h tanߠ (1.30) dx = h sec2 ߠ dߠ (1.31) cosߠ = h/r r = h/ cosߠ = h secߠ (1.32) Substituting eqs. (1.30), (1.31) and (1.32) in (1.29) we get, dE = σ ( ୦ ୲ୟ୬θ)(୦ ୱୣୡమθ ୢθ) ଶఌ୦ ୱୣୡθమ cosߠ az dE = σ ଶఌ sinߠ dߠ az On integrating ࡱ = ߪ 2ߝ න ߠ݊݅ݏ ఈ ݀ߠ ࢇࢠ ࡱ = ߪ 2ߝ ሾ−ܿߠݏሿ ఈ ࢇࢠ ࡱ = ߪ 2ߝ (1 − ܿࢇ)ߙݏࢠ (1.33) From figure 1.7 ܿߙݏ = ℎ ඥ(ܽଶ + ℎଶ)
- 13. ∴ ࡱ = ߪ 2ߝ ቆ1 − ℎ ඥ(ܽଶ + ℎଶ) ቇ ࢇࢠ ܰ ܥ (1.34) For infinite disc, radius ‘a’ tends to infinite and α = 90. ࡱ = ߪ 2ߝ (1 − ܿࢇ)09 ݏࢠ ࡱ = ߪ 2ߝ ࢇࢠ ܰ ܥ ݎ ܸ ݉ (1.35) From equation (1.35), it can be seen that electric field due to infinite disc is independent of distance. Electric field is uniform. Problems 1. Determine the magnitude of electric field at a point 3 cm away from the mid point of two charges of 10-8 c and -10-8 c kept 8 cm apart as shown in figure. Ans. Er = 2 |E| cos ߠ |E1|=|E2|=E ܧ = 1 4ߨߝ ܳ ܴଶ ܧ = 1 4ߨ 8.854 × 10ିଵଶ 10ି଼ 0.05ଶ = 36 ݉/ܸ ܭ Er = 2 36 × 103 cos 53.1 = 57.6 K V 2. A uniform charge distribution infinite in extent lies along z-axis with a charge density of 20 nc/m. Determine electric field intensity at (6,8,0). Ans. d = (6-0)i+(8-0)j+0 k = 6i + 8j d= ඥ(6ଶ + 8ଶ) = 10 ࡱ = ߣ 2ߨߝ݀ ࢇࢊ ܰ ܥ ࡱ = 20 × 10ିଽ 2ߨ 8.854 × 10ିଵଶ 10 6i + 8j 10 E = 21.57i + 28.76j N/C 3. A straight wire has an uniformly distributed charge of 0.3×10-4 c/m and of length 12 cm. Determine the electric field at a distance of 3 cm below the wire and displaced 3 cm to the right beyond one end as shown in figure. Ans. tan α1 = = .ଷ .ଵଶା.ଷ = .ଷ .ଵହ
- 14. α1 = tan-1 (0.03/0.15)= 11.3o tan β = 0.03/0.03 = 1 β = 45o , α2 = 180-45= 135o ࡱ = ߣ 4ߨߝℎ ൣ(ߙ݊݅ݏଶ − ߙ݊݅ݏଵ)ࢇ࢞ + (ܿߙݏଶ + ܿߙݏଵ) ࢇ࢟൧ ࡱ = 9 × 10ଽ × 0.3 × 10ିସ 0.03 ൣ(sin 135 − sin 11.3)ࢇ࢞ + (cos 135 + cos 11.3) ࢇ࢟൧ ࡱ = 4.59 × 10 ࢇ࢞ + 2.4 × 10 ࢇ࢟ ࡱ = ൫4.59 ࢇ࢞ + 2.4 ࢇ࢟൯ ݉/ݒܭ 4. Find the force on a point charge 50 µc at (0,0,5) m due to a charge 0f 500π µc. It is uniformly distributed over a circular disc of radius 5 m. Ans. ߪ = ொ గ మ ߪ = 500π 10ି ߨ 5ଶ = 20 µ c/m ࡱ = ߪ 2ߝ (1 − ܿࢇ)ߙݏࢠ ࡱ = 20 × 10ି 2 × 8.854 × 10ିଵଶ (1 − ܿࢇ )54 ݏࢠ ࡱ = 3.3 × 10ହ ࢇࢠ ܰ/ ܥ F= E q F = 3.3 × 10ହ × 50 × 10ି = 16.5 ܰ F = 16.5 az N 5. A charge of 40/3 nc is distributed around a ring of radius 2m, find the electric field at a point (0,0,5)m from the plane of the ring kept in Z=0 plane.
- 15. Ans. Q = 40/3 n c ߣ = ܳ 2 ߨ ܽ = 40 3 × 10ିଽ 2 ߨ 2 = 1.06 × 10ିଽ ܿ ݉ ࡱ = ߣ ܽ ℎ 2ߝݎଷ ࢇࢠ N C ࡱ = 1.06 × 10ିଽ × 2 × 5 2 × 8.854 × 10ିଵଶ × √29 ଷ ࢇࢠ ࡱ = 3.83 ࢇࢠ ܸ 6. An infinite plane at y=3 m contains a uniform charge distribution of density 10-8 /6π C/m2 . Determine electric field at all points. Ans. (i) y>3 E = σ/2ϵ j = 10-8 /(2×6π×8.854×10-12 ) j = 29.959 j Magnitude of electric field at all points is same since the electric field due to an infinite plane (or) disc is uniform. For y<3 E = -29.959 j For y<3, electric field is directed along negative y-axis. Unit vector along negative y- direction is –j. 7. A straight line of charge of length 12 cm carries a uniformly distributed charge of 0.3×10-6 coulombs per cm length. Determine the magnitude and direction of the electric field intensity at a point (i) Located at a distance of a 3 cm above the wire displaced 3 cm to the right of and beyond one end. (ii) Located at the distance of 3 cm from one end, in alignment with, but beyond the wire. (iii) Located at a distance of 3 cm from one end, on the wire itself. ans. (i) tan α1 = = .ଷ .ଵଶା.ଷ = .ଷ .ଵହ α1 = tan-1 (0.03/0.15)= 11.3o tan β = 0.03/0.03 = 1 β = 45o , α2 = 180-45= 135o
- 16. ࡱ = ߣ 4ߨߝℎ ൣ(ߙ݊݅ݏଶ − ߙ݊݅ݏଵ)ࢇ࢞ + (ܿߙݏଶ + ܿߙݏଵ) ࢇ࢟൧ ࡱ = 9 × 10ଽ × 0.3 × 10ିସ 0.03 ൣ(sin 135 − sin 11.3)ࢇ࢞ + (cos 135 + cos 11.3) ࢇ࢟൧ ࡱ = 4.59 × 10 ࢇ࢞ + 2.4 × 10 ࢇ࢟ ࡱ = ൫4.59 ࢇ࢞ + 2.4 ࢇ࢟൯ ݉/ݒܭ (ii) consider an element dx of charge λdx at a distance x from A. The field at P due to the positive charge element λdx is directed to the right. ࢊࡱ = ߣ݀ݔ 4ߨߝ(ܮ + ℎ − )ݔଶ ࢇ࢞ ࡱ = ߣ 4ߨߝ න ݀ݔ (ܮ + ℎ − )ݔଶ ࡸ ࢇ࢞ ࡱ = ߣ 4ߨߝ 1 ℎ − 1 ܮ + ℎ ൨ ࢇ࢞ ࡱ = 9 × 10ଽ × 0.3 × 10ିସ 1 0.03 − 1 0.15 ൨ ࢇ࢞ ࡱ = 7.2 × 10 ࢇ࢞ ࢜/ (iii) Now L=AC= 6 cm = 0.06 cm h = 0.03 m ࡱ = ߣ 4ߨߝ 1 ℎ − 1 ܮ + ℎ ൨ ࢇ࢞ ࡱ = 9 × 10ଽ × 0.3 × 10ିସ 1 0.03 − 1 0.09 ൨ ࢇ࢞
- 17. E = 60 ax K v/cm 1.4) Work Done: If a point charge ‘Q’ is kept in an electric field it experience a force F in the direction of electric field. Fa is the applied force in a direction opposite to that of F. Let dw be the work done in moving this charge Q by a distance dl m. Total work done in moving the point charge from ‘a’ to ‘b’ can be obtained by integration. ܹ = න ݀ ݓ ܹ = න ࡲࢇ ∙ ࢊ ܹ = − න ࡲ ∙ ࢊ (1.36) E = F/Q F = E Q ܹ = − න ࡱ ܳ ∙ ࢊ ܹ = −ܳ න ࡱ ∙ ࢊ (1.37) E = Ex ax + Ey ay + Ez az dl = dx ax + dyay + dz az E ∙ dl = EX dx + Ey dy + Ez dz ∴ ܹ = −ܳ න (E୶ dx + E୷ dy + E dz) (௫మ,௬మ,௭మ) (௫భ,௬భ,௭భ) (1.38) Problems
- 18. 1. Find the work involved in moving a charge of 1 c from (6,8,-10) to (3,4,-5) in the field E = -x i +y j –3 k. Ans. ܹ = −ܳ න (E୶ dx + E୷ dy + E dz) (ଷ.ସ.ିହ) (,଼,ିଵ) ܹ = −1 ቈන −ݔ݀ ݔ + ଷ න ݕ݀ ݕ ସ ଼ − න ݖ݀ ݖ ିହ ିଵ ܹ = − ቆ −ݔଶ 2 ቇ ଷ + ቆ ݕଶ 2 ቇ ଼ ସ − (3)ݖିଵ ିହ ൩ ܹ = − −9 2 + 36 2 + 16 2 − 64 2 − −15 + 30൨ W= 25.5 J 2. Find the work done in moving a point charge of -20 µc from the origin to (4,0,0) in the field E = (x/2 + 2y) i + 2x j Ans. ܹ = −ܳ න (E୶ dx + E୷ dy + E dz) (௫మ,௬మ,௭మ) (௫భ,௬భ,௭భ) dl = d xi ܹ = −ܳ න (E୶ dx ) (ସ,,) (,,) ܹ = 20 × 10ି න ((x/2 + 2y) dx ) (ସ,,) (,,) ܹ = 20 × 10ି ቆ ݔଶ 4 ቇ ସ W = 80×10-6 J 1.5) Absolute Potential: Absolute potential is defined as the work done in moving a unit positive charge from infinite to the point against the electric field.
- 19. A point charge Q is kept at an origin as shown in figure. It is required to find the potential at ‘b’ which is at distance ‘r’ m from the reference. Consider a point R1 at a distance ‘x’ m. The small work done to move the charge from R1 to P1 is dw. The electrical field due to a point charge Q at a distance ‘x’ mis ࡱ = ܳ/(4ߨߝݔଶ ) Work done = E• dx Total work done can be obtained by integration Work done (W) = − Q ۳ ∙ dܔ ୠ ୟ V= − 1 ۳ ∙ dܔ ୠ ୟ V = − න ۳ ∙ dܔ ୠ ୟ E = Ex ax , dl = dx ax V = − න ܳ 4ߨߝݔଶ ݀ݔ ୠ ୟ ܸ = ܳ 4ߨߝݎ (1.39) 1.6) Potential Difference: Vab Potential difference Vab is defined as the work done in moving a unit positive charge from ‘b’ to ‘a’.
- 20. Consider a point charge Q kept at the origin of a spherical co-ordinate system. The field is always in the direction of ar. No field in the direction of ߠ and ϕ. The points ‘a’ and ’b’ are at distance ra and rb respectively as shown in figure. Vୟୠ = − න ۳ ∙ dܔ ୟ ୠ E = Er ar , dl = dr ar Vୟୠ = − න E୰d୰ ୟ ୠ Vୟୠ = − න ܳ 4ߨߝݎଶ d୰ ୰ ୰ౘ Vୟୠ = ܳ 4ߨߝ 1 ݎ − 1 ݎ ൨ Vୟୠ = ܳ 4ߨߝ 1 ݎ − ܳ 4ߨߝ 1 ݎ Vୟୠ = Vୟ − ܸ (1.40) 1.6.1) Potential difference due to line charge: The wire is uniformly charged with λ C/m. We have to find the potential difference Vab due to this line charge. Consider a point P at a distance P from the line charge.
- 21. Fig. Line charge E = Eρ aρ dl = dρ aρ E•dl = Eρ aρ • dρ aρ = Eρ dρ Potential difference Vab is the work done in moving a unit +ve charge from ‘b’ to ‘a’. Vୟୠ = − න ۳ ∙ dܔ ୟ ୠ Vୟୠ = − න Eρ dρ ୟ ୠ Vୟୠ = − න ߣ 2ߨߝߩ dρ ρ ρౘ Vୟୠ = ߣ 2ߨߝ ln ߩ ߩ (1.41) 1.6.2) Potential due to charged ring: A thin wire is bent in the form of a circular ring as shown in figure. It is uniformly charged with a charge density λ C/m. It is required to determine the potential at height ‘h’ m from the centre of the ring. The ring is assumed to be formed by several point charges. Fig. Charged ring Let dv be the potential due to the line charge element of length dl containing a charge dQ. ݀ݒ = ݀ܳ 4ߨߝݎ ݀ݒ = ߣ݈݀ 4ߨߝݎ ܸ = න ߣ݈݀ 4ߨߝݎ
- 22. ܸ = ߣ 4ߨߝݎ න ݈݀ ܸ = ߣ 4ߨߝݎ 2ߨܽ ܸ = ߣܽ 2ߝݎ ܸ = ߣܽ 2ߝ√ܽଶ + ܾଶ )24.1( ݏݐ݈ݒ 1.6.3) Potential due to a charged disc: let dv be the potential due to one ring. Each ring is assumed to be having several point charges dQ1, dQ2, ............ dQn. Potential due to the ring is the sum of potential. Fig. Charge disc ݀ݒ = ݀ܳଵ 4ߨߝݎ + ݀ܳଶ 4ߨߝݎ + … … … … … . + ݀ܳ 4ߨߝݎ ݀ݒ = ݀ܳଵ + ݀ܳଶ + … … . . + ݀ܳ 4ߨߝݎ ݀ݒ = ݀ܳ 4ߨߝݎ ݀ݒ = ߪ ݀ݏ 4ߨߝݎ = ߪ 2ߨݔ݀ ݔ 4ߨߝݎ = ߪ ݔ݀ ݔ 2ߝݎ Potential due to entire disc can be obtained by integration ܸ = න ݀ݒ = න ߪ ݔ݀ ݔ 2ߝݎ = ߪ 2ߝ න ݔ √ݔଶ + ℎଶ ݀ ݔ Let ݔଶ + ℎଶ = ݐଶ 2x dx = 2t dt Therefore we have
- 23. ܸ = ߪ 2ߝ න ݐ݀ ݐ ݐ √మାమ ܸ = ߪ 2ߝ න ݀ݐ √మାమ ܸ = ߪ 2ߝ ቀඥܽଶ + ℎଶ − ℎቁ )34.1( ݏݐ݈ݒ At the centre of the disc , h=0; ܸ = ߪܽ 2ߝ )44.1( ݏݐ݈ݒ Problems 1. Find the work done in moving a poimt charge Q= 5 µ c from origin to (2m, π/4 , π/2), spherical co-ordinates in the field E = 5 e-r/4 ar + (10/(r sin θ)) aϕ v/m. Ans. dl = dr ar + rdθ aθ + r sinθ dϕ aϕ E• dl = 5 e-r/4 dr + 10 dϕ W = − Q න ۳ ∙ dܔ ୠ ୟ W = − 5 × 10ି ቈන 5 eି୰/ସ dr ଶ න 10 dϕ π/ଶ W= -117.9 µ J. 2. 5 equal, point charges of 20 nc are located at x=2,3,4,5 and 6 cm. Determine the potential at the origin. Ans. By superposition principal V= V1 + V2 + V3 + V4 + V5 ܸ = ܳ 4ߨߝݎଵ + ܳ 4ߨߝݎଶ + ܳ 4ߨߝݎଷ + ܳ 4ߨߝݎସ + ܳ 4ߨߝݎହ
- 24. ܸ = 20 × 10ି 4ߨߝ ൬ 1 2 + 1 3 + 1 4 + 1 5 + 1 6 ൰ = 261 ݏݐ݈ݒ 3. A line charge of 0.5 nc/m lies along z-axis. Find Vab if a is (2,0,0) , and b is (4,0,0). Ans. Vୟୠ = ߣ 2ߨߝ ln ߩ ߩ Vୟୠ = 0.5 × 10ିଽ 2ߨ × 8.854 × 10ିଵଶ ln 4 2 Vab = 6.24 V. 4. A point charge of 0.4 nc is located at (2,3,3) in Cartesian coordinates. Find Vab if a is (2,2,3) and b is (-2,3,3). Ans. Vୟୠ = Vୟ − ܸ Vୟୠ = ܳ 4ߨߝ 1 ݎ − ܳ 4ߨߝ 1 ݎ ra = (2-2) i + (2-3) j + (3-3) k = -j ra = 1 rb = (-2-2) i + (3-3) j + (3-3) k = -4 i rb = 4 Vୟୠ = 0.4 × 10ିଽ 4ߨ × 8.854 × 10ିଵଶ 1 1 − 1 4 ൨ = 2.7 ݒ 5. A total charge of 40/3 nc is distributed around a ring of radius 2m. Find the potential at a point on the z-axis 5m from the plane of the ring kept at z=0. What would be the potential if the entire charge is concentrated at the centre. ans. ܸ = ߣܽ 2ߝ√ܽଶ + ܾଶ ݏݐ݈ݒ ߣ = ܳ ݈ = 40 3 × 10ିଽ 2ߨ 2 = 1.06 × 10ିଽ ܥ ݉
- 25. ܸ = 1.06 × 10ିଽ × 2 2 × 8.854 × 10ିଵଶ × √2ଶ + 5ଶ = 22.2 ݒ 6. Determine the potential at (0,0,5)m due to a total charge of 10-8 c distributed uniformly along a disc of radius 5m lying in the Z=0 plane and centered at the orign. Ans. ܸ = ߪ 2ߝ ቀඥܽଶ + ℎଶ − ℎቁ ݏݐ݈ݒ ߪ = ܳ ܽܽ݁ݎ = 10ି଼ ߨ × 5ଶ ܸ = 10ି଼ ߨ × 5ଶ 2ߝ ቀඥ5ଶ + 5ଶ − 5ቁ V= 14.89 v 1.7) Relation between V and E: Consider a point charge Q at the origin as shown in figure. Electric field due to this charge at the point ‘P’ is ࡱ = ܳ 4ߨߝݎଶ ࢇ࢘ Consider સ ൬ 1 ݎ ൰ = ൬ ߲ ߲ݎ ࢇ࢘ + … … … ൰ ൬ 1 ݎ ൰ સ ൬ 1 ݎ ൰ = − ൬ 1 ݎଶ ൰ ࢇ࢘ Using above expression in E , we get ࡱ = ܳ 4ߨߝ − સ ൬ 1 ݎ ൰ ࡱ = −સ ܳ 4ߨߝ ൬ 1 ݎ ൰ Therefore, ࡱ = −સ ܸ
- 26. Problem 1. Given V= (x-2)2 × (y+2)2 ×(z-1)3 . Find electric field at the origin. Ans. V= (x-2)2 × (y+2)2 ×(z-1)3 ࡱ = −સ ܸ ࡱ = − ൬ ∂v ∂x i + ∂v ∂y j + ∂v ∂z k൰ ࡱ = −2(x − 2)× (y+2)2 ×(z-1)3 i-(x-2)2 × 2(y+2) ×(z-1)3 j - (x-2)2 × (y+2)2 ×3(z-1)2 k Electric field at the origin ࡱ = −2(0 − 2) (0+2)2 ×(0-1)3 i -(0-2)2 × 2(0+2) ×(0-1)3 j - (0-2)2 × (0+2)2 ×3(0-1)2 k E = -(16i+16j+48k) v/m. 1.8) Poisson’s and Laplace’s Equations: From the Gauss law we know that .ܦ ݀ݏ = Q (1) A body containing a charge density ρ uniformly distributed over the body. Then charge of that body is given by Q = ૉ ݀ݒ (2) .ܦ ݀ݏ = ૉ ݀ݒ (3) This is integral form of Gauss law. As per the divergence theorem .ܦ ݀ ݏ = સ. ۲ ݀ݒ (4) સ. ۲ = ૉ (5) This is known as point form or vector form or polar form. This is also known as Maxwell’s first equation. D = ƐE (6) સ. ƐE = ρ સ. E = ࣋/Ɛ (7) We know that E is negative potential medium
- 27. E = -સ܄ (8) From equations 7 and 8 સ. (– સ)܄ = ࣋/Ɛ સ2 V = - ࣋/Ɛ (9) Which is known as Poisson’s equation in static electric field. Consider a charge free region (insulator) the value of ρ = 0, since there is no free charges in dielectrics or insulators. સ2 V = 0 This is known as Laplace’s equation.