Question

191-2e 92-4e •d Two charges, 91 and 22 lie a distance d = 3 cm apart on the x-axis. Charge qz is located at the origin and qz is located at x + 3 cm. What is the electric potential at the point, in question 6, where E is zero? A. 56 x 10v O B. 28 x 10V C.0.0V D. -56 x 10 V Reset Selection

Answer 1

Let P be the point between the charges, where the electric field is zero.

Let P be at a distance of  
d cm
from $q_1$ and therefore at a distance of
3-d cm
from  $q_2$

therefore,

at P, magnitudes of both electric field are same and they cancel out each other as they are in opposite directions

ki kq2 d2 (3 – d)2 2e 4e d2 (3 - d)2 2* (3 – d)2 = 4d 2* (9 - 6d+ d) = 4d 18 – 12d + 2d = 4d 2d2 + 12d - 18 = 0 d? + 6d - 9=0

from this quadratic equation, we get

$\\d = 1.24 \ cm \\d = 0.0124 \ m$

Therefore, the point P where electric field is zero is

At this point, the electric potential is

$\\V = \frac{kq_1}{d} + \frac{kq_2}{3 - d} \\\\V = k * (\frac{2e}{d} + \frac{4e}{3 - d}) \\\\V = k * e * (\frac{2}{0.0124} + \frac{4}{0.03 - 0.0124}) \\\\V = 9 * 10^9 * 1.602 * 10^{-19} * (161.29 + 227.27) \\\\V = 14.418 * 10^{-10} * 388.56 \\\\V = 5602.2 * 10^{-10} \ V \\\\V \approx 56 * 10^{-8} \ V$