Question

An infinite thin non-conducting sheet has uniform surface charge density $\sigma $ . Find the potential difference $V_{A}-V_{B}$ between points $A$ and $B$ as shown in the figure. The line $AB$ makes an angle of $37^\circ $ with the normal to the sheet. $\left(sin 3 7 ^\circ = \frac{3}{5}\right)$

• A. $\frac{2 \sigma d}{5 \epsilon _{0}}$
• B. $\frac{- 2 \sigma d}{5 \epsilon _{0}}$
• C. $\frac{3}{10}\frac{\sigma d}{\epsilon _{0}}$
• D. $\frac{- 3 \sigma d}{10 \epsilon _{0}}$

Answer 1

Answer: (A)

$V_{f}-V_{i}=-\overset{ \rightarrow }{E}.\overset{ \rightarrow }{S}$ (for uniform field)
$V_{B}-V_{A}=-\frac{\sigma }{2 \epsilon _{0}}dcos 37^\circ =\frac{- 2 \sigma d}{5 \epsilon _{0}}$
$\Rightarrow V_{A}-V_{B}=\frac{2 \sigma d}{5 \epsilon _{0}}$