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# Tell me? Within a spherical charge distribution of charge density ρ(r), N equipotential surfaces of potential V0, V0 + ∆V, V0 + 2∆V, .......... V0 + N∆V (∆V>0), are drawn and have increasing radii r0, r1, r2,..........rN, respectively. If the

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)

• Option 2)

• Option 3)

• Option 4)

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N

As we learnt in

Relation between field and potential -

$\dpi{100} E=\frac{-dv}{dr}$

- wherein

$\dpi{100} \frac{dv}{dr} -$   Potential gradient.

If P at the surface r = R -

$\dpi{100} E_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R^{2}}$       $\dpi{100} V_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}$

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

-

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)

Incorrect option

Option 2)

Incorrect option

Option 3)

Correct option

Option 4)

Incorrect option

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