Consider a block of conducting material ofresistivity '$\rho$' shown in the figure. Current '$I$' enters at '$A$' and leaves from '$D$'. We apply superp osition principle to find voltage '$\Delta V$ ' developed between '$B$' and '$C$'. The calculation is done in the following steps:
$(i)$ Take current '$I$' entering from '$A$' and assume it to spread over a hemispherical surface in the block.
$(ii)$ Calculatefield $E(r)$ at distance '$r$' from $A$ by using Ohm's law $E = \rho j$, where j is the current per unit area at '$r$'.
$(iii)$ From the '$r$' dependence of $E(r)$, obtain the potential $V(r)$ at $r$.
$(iv)$ Repeat $(i), (ii)$ and $(iii)$ for current '$I$' leaving '$D$' and superpose results for '$A$' and '$D$'.
$(i)$ Take current '$I$' entering from '$A$' and assume it to spread over a hemispherical surface in the block.
$(ii)$ Calculatefield $E(r)$ at distance '$r$' from $A$ by using Ohm's law $E = \rho j$, where j is the current per unit area at '$r$'.
$(iii)$ From the '$r$' dependence of $E(r)$, obtain the potential $V(r)$ at $r$.
$(iv)$ Repeat $(i), (ii)$ and $(iii)$ for current '$I$' leaving '$D$' and superpose results for '$A$' and '$D$'.
For current entering at $A$, the electric field at a distance '$r$'
from $A$ is
AIEEE 2008, Diffcult
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Let j be the current density.
Then $j \times 2 \pi r^{2}=I \Rightarrow j=\frac{I}{2 \pi r^{2}} \therefore E=\rho j=\frac{\rho I}{2 \pi r^{2}}$
Now, $\Delta V_{BC}^\prime = - \int\limits_{a + b}^a {\vec E} .\overline {dr} = - \int\limits_{a + b}^a {\frac{{\rho I}}{{2\pi {r^2}}}dr} $
$=-\frac{\rho I}{2 \pi}\left[-\frac{1}{r}\right]_{a+b}^{a}=\frac{\rho I}{2 \pi a}-\frac{\rho I}{2 \pi(a+b)}$
On applying superposition as mentioned we get
$\Delta V_{B C}=2 \times \Delta V_{B C}^{\prime}=\frac{\rho I}{\pi a}-\frac{\rho I}{\pi(a+b)}$
As shown above $\quad E=\frac{\rho I}{2 \pi r^{2}}$
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