In the following right angle AB 2 m and angle ACB is 6343 D

In the following right angle, AB = 2 m, and angle ACB is 63.43 Degree. A point charge of 2* 28 nC is placed at point A and another point charge -3* 28 nC is placed at point C. Calculate the POTENTIAL at B.

Solution

AB = 2m

by trigonometry

tan theta = AB/BC

BC = AB/tan63.43

BC = 2/tan63.43

BC = 1.002m

V = kq1/r1 + kq2/r2

V = k(2 * 28 * 10^-9 / 2 - 3*28 * 10^-9 /1.002)

V = 28 * 9 * 10^9 * 10^-9 ( 2/2 - 3/1.002)

V = -503.8 V