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Within a spherical charge distribution of charge density ρ(r), N equipotential surfaces of potential V₀, V₀+ V, V₀+ 2V, .......... V₀ + NV (V>0), are drawn and have increasing radii r₀, r₁, r₂,..........r_N, 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)

best_answer

As we learnt in

Relation between field and potential -

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

- 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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Plabita

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