Question

A -10.0 nC point charge and a +20.0nC point charge are 13.5 cm apart on the x-axis.

Part A

What is the electric potential at the point on the x-axis where the electric field is zero?

Express your answer with the appropriate units.

Part B

What is the magnitude of the electric field at the point on the x-axis, between the charges, where the electric potential is zero?

Express your answer with the appropriate units.

Answer 1

(a) When the points have different charges and magnitudes, the zero-field point (and there can be only one) lies outside the charge of smaller magnitude. This is because between them, the field points toward the negative charge, and outside the larger charge, the larger charge always dominates because of proximity. Hence if we're measuring "x" from the smaller charge,

E = kQ / d² = k * 10nC / x² = k * 20nC / (x + 0.135m)²

10(x + 0.135)² = 20x² → for x in meters.

x = 0.326m

At this point, then,

potential V = kQ / d = 9e9 N·m²/C² * (20/(0.135+0.326) - 10/0.326)nC/m

V = 114.38 V

(b) Electric potential from a charge is K*q/r. If we consider the -10nC charge at the origin, then the potential from it is -K*10/|x|; The potential from the 17nC charge is +K*20/|x-13.5|. They must sum to zero, so

20/|x-13.5| - 10/|x| = 0,

=> 20*|x| = 10*|x-13.5|

=>2*|x| = |x – 13.5|

2x = x – 13.5 or 2x = 13.5 - x

x = -13.5 or x = 4.5cm. Plug these in and you'll see they both result in 0 potential at those points.

Only the value of x = 4.5cm is between the charges. The field from the -10nC charge is toward the origin, and the field from the 20nC charge is also toward the origin. The fields will then sum directly

The electric field is

E = [10*10^-9/(0.045)^2 + 20*10^-9/(0.135 – 0.045)^2]*9*10^9

E = 5.00*10^4 N/C

The field points to the -8.5nC charge