Question

Two point charges of +2 μC and -8 μC are located on the x-axis at x = -1 cm and x = +2 cm respectively. Where should a third charge of +46 μC be placed on the +x-axis so that the potential at the origin is equal to zero? (in units of cm)

Answer 1

Electric potential is a scalar quantity which is given by:

V = k*Q/R

Now given that net electric potential due to these three charges at origin is zero, So

V_net = V1 + V2 + V3 = 0

V_net = k*q1/R1 + k*q2/R2 + k*q3/R3 = 0

k*q1/R1 + k*q2/R2 = -k*q3/R3

q1/R1 + q2/R2 = -q3/R3

R3 = -q3/(q1/R1 + q2/R2)

Given that q1 = 2*10^-6 C at x = -1 cm = -0.01 m

R1 = distance between q1 and origin = 0.01 m

q2 = -8*10^-6 C at x = 2 cm = 0.02 m

R2 = distance between q2 and origin = 0.02 m

And q3 = 46*10^-6 C at suppose x = 'd',

R3 = d

So Using given values:

d = -46*10^-6/(2*10^-6/0.01 - 8*10^-6/0.02)

d = -46/(2/0.01 - 8/0.02) = 0.23 m

x = 0.23 m = 23 cm

So third charge should be at x = 23 cm

Let me know if you've any query.