Question

A 13.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 4 cm .

Part A

At what point or points on the x-axis is the electric potential zero?

Express your answer using two significant figures. If there is more than one answer, give each answer separated by a comma.

Answer 1

The electric potential due to multiple charges is:

V = kQ1/r1 + kQ2/r2

where k = 1/(4*π*ε) ~ 9*10^9

Let "x" be the location of a point being tested for potential:

If between the two charges:

V = k*Q1/(x - 0) + k*Q2/(.04 - x) = 0

The "k" cancels:

Q1*(.04 - x) + Q2*x = 0

x*(Q2 - Q1) = -.04*Q1

x = -.04*Q1/(Q2 - Q1) = -.04*13.5*10^-9/(-1.2*10^-9 - 13.5*10^-9) = .036 m, or 3.6 cm

If below the origin, using "-x" keeps the first radius value positive :

V/k = Q1/(-x) + Q2/(.04 - x) = 0

Q1*(.04 - x) + Q2*(-x) = 0

x(-Q1 - Q2) = -.04*Q1

x = .04*Q1/(Q1 + Q2) = .043m, but this is not an "x" value on the negative axis, so the only true zero on the negative axis is x = -∞. When "r" approaches infinity, the denominator radius drives all of the contributing potentials to zero.

Above 0.04 on the positive x-axis:

V/k = Q1/(x) + Q2/(x - .04) = 0

Q1*(x - .04) + Q2*x = 0

x*(Q1 + Q2) = .04*Q1

x = .04*Q1/(Q1 + Q2) = .04*13.5*10^-9/(13.5*10^-9 - 1.2*10^-9) = .043 m or 4.3 cm.