Question

Two charges are set on the x-axis 15.5 cm away from each other. The charges are -7.70 nC and 20.4 nC. Calculate the electric potential at the point on the x-axis where the electric field due to these two charges is zero.

Answer 1

Assuming Q1 (-7.70 nC) is at x = 15.5 cm (0.155 m) & Q2 (20.4 nC) is at x = 0 cm

Now suppose at distance d from charge -7.70 nC net electric field is zero, then

Electric field is given by:

E = kQ/R^2

Since we know that direction of electric field due to +ve charge will be away from charge and due to -ve charge will be towards the charge. In given case since Q1 is -ve and Q2 is +ve, So electric field will be zero towards the right side of Q1 (Since |Q2| > |Q1|)

Enet = E1 - E2 = 0

E1 = E2

kQ1/r1^2 = kQ2/r2^2

r1 = d

r2 = d + 0.155

Using above values:

Q1/r1^2 = Q2/r2^2

7.70/d^2 = 20.4/(d + 0.155)^2

(d + 0.155)/d = sqrt (20.4/7.70) = 1.6277

d + 0.155 = 1.6277*d

d = 0.155/(1.6277 - 1)

d = 0.247 m = 24.7 cm

So electric field will be zero at d = 24.7 cm (see that this d is not from origin but from charge q1)

Now at this location electric potential due to both charge will be:

electric potential is an scalar quantity which is given by:

V_net = k*Q1/r1 + k*Q2/r2

r1 = d = 0.247 m

r2 = d + 0.155 = 0.247 + 0.155 = 0.402

V_net = 9*10^9*(-7.70*10^-9)/0.247 + 9*10^9*20.4*10^-9/0.402

V_net = 176 V