# For current entering at A, the electric field at a distance r  from A  is Directions : Consider a block of conducting material of resistivity  $\rho$ shown in the figure. Current I enters at A    and leaves from  D  . We apply superposition 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)     Calculate field 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 B Option 1) $\frac{\rho I}{4\pi r^{2}}$ Option 2) $\frac{\rho I}{8\pi r^{2}}$ Option 3) $\frac{\rho I}{ r^{2}}$ Option 4) $\frac{\rho I}{2\pi r^{2}}$

As we learnt in

If current density is uniform -

$\vec{J}=\frac{\vec{I}}{A}$

- wherein

$\vec{J-}$ Current density

$\vec{A-}$ Normal cross-section

Current density, $j= \frac{I}{2\pi r^{2}}$'

$Resistance= \frac{\rho I}{area}= \frac{\rho r}{2\pi r^{2}}$

$E= \frac{I\rho }{2\pi r^{2}}$

Option 1)

$\frac{\rho I}{4\pi r^{2}}$

This option is incorrect.

Option 2)

$\frac{\rho I}{8\pi r^{2}}$

This option is incorrect.

Option 3)

$\frac{\rho I}{ r^{2}}$

This option is incorrect.

Option 4)

$\frac{\rho I}{2\pi r^{2}}$

This option is correct.

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