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Within a spherical charge distribution of charge density  ρ(r), N equipotential surfaces of potential V0, V+ V, V0 + 2V, .......... V0 + NV  (V>0), are drawn and have increasing radii r0, r1, r2,..........rN, respectively.  If the difference in the radii of the surfaces is constant for all values of V0 and V then :

  • Option 1)

     \rho (r)\alpha r

  • Option 2)

    \rho (r)= constant

  • Option 3)

    \rho (r)\alpha \frac{1}{r}

  • Option 4)

    \rho (r)\alpha \frac{1}{r^{2}}


Answers (2)


As we learnt in 

Relation between field and potential -


- wherein

\frac{dv}{dr} -   Potential gradient.



If P at the surface r = R -

E_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R^{2}}       V_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}

\dpi{100} E_{s}=\frac{\rho R}{3 \epsilon _{0}}               V_{s}=\frac{\rho R^{2}}{3 \epsilon _{0}}



 We know E= -\frac{\partial v}{\partial r}

Here \Delta v and \Delta r are same for any pair of surfaces.

E= constant

Now, electric field inside the spherical charge distribution 

E= \frac{\rho }{3\epsilon _0}r

E would be constant of \rho r= constant

\rho (r) \propto\frac{1}{r}



Option 1)

 \rho (r)\alpha r

Incorrect option

Option 2)

\rho (r)= constant

Incorrect option

Option 3)

\rho (r)\alpha \frac{1}{r}

Correct option

Option 4)

\rho (r)\alpha \frac{1}{r^{2}}

Incorrect option

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