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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  . 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}}


Answers (1)


As we learnt in 

If current density is uniform -


- 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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