Question

When an electron moves from A to B along an electric field line in Fig. 24-34, the electric field does 3.94 x 10^-19 Jof work on it. What are the electric potential differences (a) V_B - V_A, (b) V_C - V_A, and (c) V_C - V_B?

Short answer

1. The electric potential difference is V_B - V_A = 2.46 V.
2. The electric potential difference is V_C - V_A = 2.46 V.
3. The electric potential difference is V_C - V_B = 0.

Step-by-step solution

Given data:

The electric field energy is 3.94 x 10^-19 J.

Understanding the concept:

Using the equation of potential energy, you can find the electric potential difference.

$\mathbf{∆}\mathit{U}\mathbf{=}\mathit{e}\mathbf{∆}\mathit{V}$

Here, $\mathbf{∆}\mathit{U}$ is the change in potential energy, e is the electric charge having a value 1.60 x 10^-19 C, and $\mathbf{∆}\mathit{V}$is the electric potential difference.

(a) Calculate VB - VA :

The electric potential difference between A and B is define as below.

$\begin{array}{rcl}{V}_B-V_A& =& \frac{∆U}{q}\\ & =& \frac{-W}{\left(-e\right)}\\ & =& \frac{-\left(3.94\times {10}^{-19}J\right)}{\left(-1.60\times {10}^{-19}C\right)}\\ & =& 2.46V\end{array}$

(b) Calculate VC - VA:

The electric potential difference between A and C is define as below.

$\begin{array}{rcl}{V}_C-V_A& =& V_B-V_A\\ & =& 2.46V\end{array}$

(c) Calculate VC - VB :

Since C and B are on the same equipotential line, the potential difference between C and B is,

V_C - V_B = 0