Question

Two point charges, one of -1.5 μC and the other of +3.7 μC, are placed at the points (-2.5 m, -1.5 m) and (1.2 m, 0), respectively. (a) Find the electric potential at the origin. (b) There is one point along the radial line connecting the two charges where the electric potential is zero. Find this point (x- and y-coordinates).

**Plase guide me through**

Answer 1

a) The distance from -1.5uC to the origin is r1 = sqrt(2.5^2+1.5^2) = 2.92 m

the distance from +3.7uC to the origin is r2 = 1.2 m

potential due to -1.5uc is V1 = k*q1/r1 = (-9*10^9*1.5*10^-6)/(2.92) = -4623.28 V

Potential due to +3.7uC is V2 = k*q2/r2 = (9*10^9*3.7*10^-6)/1.2 = 27750 V

then Net potential at origin is V = V1+V2= -4623.28+27750 = 23126.72 V

b) distance between two chearges is d = sqrt((1.2+2.5)^2+1.5^2) = 4 m

Given that potential is zero

V1 = v2

k*Q1/r1 = k*Q2/r2

Q1/d = Q2/(4-d)

1.5/d = 3.7/(4-d)

d = 1.15 m

direction is theta = tan^(-1)(1.5/3.7) = 22.3deg

the

required x-coordinate is = -(4-1.15)*cos(22.3) + 1.2 = 1.44 m

y-coordinate is -(4-1.15)*sin(22.3) = -1.1 m