Question

A current of 0.01 mA passes through the potentiometer wire of a resistivity of ${10}^9\Omega$ cm and area of cross-section ${10}^{-2}c{m}^2$. The potential gradient is

• A. ${10}^{9}V/m$
• B. ${10}^{11}V/m$
• C. ${10}^{10}V/m$
• D. ${10}^{8}V/m$

Answer 1

Answer: (D)

${10}^{8}V/m$

Potential gradient is
$k=\frac{V}{l}=\frac{iR}{l}$ $(\therefore V=iR)$
or $k=\frac{i\times \rho \frac{l}{A}}{l}$ $(\because R=\rho \frac{l}{A})$
$⟹k=\frac{i\rho }{A}$

where $i=0.01\times {10}^{-3}A$, $\rho ={10}^9\times {10}^{-2}m$ and $A={10}^{-2}\times {10}^{-4}{m}^2$
$\therefore k=\frac{0.01\times {10}^{-3}\times {10}^9\times {10}^{-2}}{{10}^{-2}\times {10}^{-4}}={10}^{8}V/m$