Chapter 24: Q14P (page 710)

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

Expert verified

The potential difference is, $- 4.5 \times 10^{3} V$ .
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, $q = 1.0 μ C$ .
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

$V = \frac{q}{4 π ε_{0} R}$

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,

$V_{A} - V_{B} = \frac{q}{4 π ε_{0} r_{A}} - \frac{q}{4 π ε_{0} r_{B}}$

Substitute all the value in the above equation.

$\begin{array}{rcl} V_{A} - V_{B} & = & \frac{q}{4 π ε_{0} r_{A}} - \frac{q}{4 π ε_{0} r_{B}} \\ V_{A} - V_{B} & = & (1.0 \times 10^{- 6} C) (8.99 \times 10^{9} N . m^{2} / C^{2}) (\frac{1}{2.0 m} - \frac{1}{1.0 m}) \\ = & - 4.5 \times 10^{3} V \end{array}$