Question

Consider a point charge q = 1.0μC, point A at distance d1 = 2.160 m from q, and point B at distance d2 = 1.200 m. If these points are diametrically opposite each other, as in the lefthand figure, what is the electric potential difference VA - VB?

What is that electric potential difference if points A and B are located as in the righthand figure?

Answer 1

Use the following formula to find the electric potential.

$$ V=\frac{k q}{r} $$

Lefthand figure:

The electric potential is,

$$ \begin{aligned} V &=V_{A}-V_{B} \\ &=\frac{k q}{d_{1}}-\frac{k q}{d_{2}} \\ &=\left(9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right)\left(1 \times 10^{-6} \mathrm{C}\right)\left(\frac{1}{2.16 \mathrm{~m}}-\frac{1}{1.2 \mathrm{~m}}\right) \\ &=-3.33 \mathrm{kV} \end{aligned} $$

Righthand figure:

The electric potential is,

$$ \begin{aligned} V &=V_{A}-V_{B} \\ &=\frac{k q}{d_{1}}-\frac{k q}{d_{2}} \\ &=\left(9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right)\left(1 \times 10^{-6} \mathrm{C}\right)\left(\frac{1}{2.16 \mathrm{~m}}-\frac{1}{1.2 \mathrm{~m}}\right) \\ &=-3.33 \mathrm{kV} \end{aligned} $$