Question

The potential difference $V_{AB}$ between $A(0,0,0)$ and $B(1,1,0)$ in an electric field $\vec{E}=x\hat{i}+z\hat{k}$, is

• A. 1 V
• B. $\frac{3}{2}V$
• C. $\frac{1}{2}V$
• D. 2 V

Answer 1

Answer: (C)

$dV=-\vec{E}\cdot d\vec{r}$
$\underset{B}{\overset{A}{\int }}dV=-\underset{(1,1,0)}{\overset{(0,0,0)}{\int }}(x\hat{i}+z\hat{k})\cdot (dx\hat{i}+dy\hat{j}+dz\hat{k})$ or
$V_A-V_B=-\underset{(1,1,0)}{\overset{(0,0,0)}{\int }}(xdx+zdz)$ or
$V_A-V_B=-{\left[\frac{x^2}{2}+\frac{z^2}{2}\right]}_{(1,1,0)}^{(0,0,0)}$
$=-\left[\frac{0}{2}+\frac{0}{2}-\frac{1}{2}-\frac{0}{2}\right]=\frac{1}{2}V$