Electric field solving
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1) The document calculates the electric field components Ex and Ey resulting from two point charges q1 = 10 nc and q2 = 15 nc located at distances of 3m and 5m respectively. 2) It finds that Ey = 14.6 N/C and Ex = -4.8 N/C, with the total electric field making an angle of -72.8 degrees. 3) It notes that when a point charge is placed in a uniform electric field, it will accelerate according to Newton's law, undergoing parabolic motion similar to projectile motion under gravity.
Electric field solving
- 1. Example continued Field due to q 1 E = 10 10 N.m 2 /C 2 10 X10 -9 C/(3m) 2 = 11 N/C in the y direction. Recall E =kq/r 2 and k=8.99 x 10 9 N.m 2 /C 2 E y = 11 N/C E x = 0 Field due to q 2 5 E = 10 10 N.m 2 /C 2 15 X10 -9 C/(5m) 2 = 6 N/C at some angle Resolve into x and y components E E y =E sin C E x =E cos C Now add all components E y = 11 + 3.6 = 14.6 N/C E x = -4.8 N/C Magnitude x y q 1 =10 nc q 2 =15 nc 4 3
- 2. Example continued E y = 11 + 3.6 = 14.6 N/C E x = -4.8 N/C E Magnitude of electric field tan -1 E y /E x = atan (14.6/-4.8)= -72.8 deg x q 1 =10 nc q 2 =15 nc 4 3
- 3. Motion of point charges in electric fields When a point charge such as an electron is placed in an electric field E, it is accelerated according to Newton’s Law: a = F/m = qE/m for uniform electric fields a = F/m = mg/m = g for uniform gravitational fields If the field is uniform, we now have a projectile motion problem- constant acceleration in one direction. So we have parabolic motion just as in hitting a baseball, etc except the magnitudes of velocities and accelerations are different. Replace g by qE/m in all equations; For example, In y =1/2at 2 we get y =1/2(qE/m)t 2