Question

Consider a particle with charge q = 1.0 mC, point Aat distance d₁ = 2.0 mfrom q, and point Bat distance d₂ = 1.0 m. (a) If Aand B are diametrically opposite each other, as in Fig. 24-36a, what is the electric potential difference V_A - V_B? (b) What is that electric potential difference if Aand Bare located as in Fig. 24-36b?

Short answer

1. The potential difference is, $-4.5\times {10}^{3}V$.
2. V (r) depends only on the magnitude of r, the outcome remains the same.

Step-by-step solution

Given

• The charge on the particle is, $q=1.0\mu C$.
• The distance between point A and charge q is, d₁ = 2.0m.
• The distance between point B and charge q is, d₂ = 1.0m.

Understanding the concept

The electric potential V at the surface of a drop of charge q and radius R is given by

$\mathit{V}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{R}}$

Using this equation, we find potential differences between two points.

(a) Calculate the electric potential difference VA - VB if A and B are diametrically opposite each other

The potential difference is expressed as,

$V_A-V_B=\frac{q}{4\pi {\epsilon }_0{r}_A}-\frac{q}{4\pi {\epsilon }_0{r}_B}$

Substitute all the value in the above equation.

$\begin{array}{rcl}{V}_A-V_B& =& \frac{q}{4\pi {\epsilon }_0{r}_A}-\frac{q}{4\pi {\epsilon }_0{r}_B}\\ V_A-V_B& =& \left(1.0\times {10}^{-6}C\right)\left(8.99\times {10}^{9}N.m^2/C^2\right)\left(\frac{1}{2.0m}-\frac{1}{1.0m}\right)\\ & =& -4.5\times {10}^{3}V\end{array}$

Hence the potential difference is, $\begin{array}{rcl}& & -4.5\times {10}^{3}V\end{array}$.

(b) Calculate electric potential difference if A and B are located as in figure

Since V (r) depends only on the magnitude of r, the outcome remains the same.