JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 15)
Within a spherical charge distribution of charge density $$\rho $$(r), N equipotential surfaces of potential V0, V0 + $$\Delta $$V, V0 + 2$$\Delta $$V, .......... V0 + N$$\Delta $$V ($$\Delta $$ 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 $$\Delta $$V then :
$$\rho $$ (r) $$\alpha $$ r
$$\rho $$ (r) = constant
$$\rho $$ (r) $$\alpha $$ $${1 \over r}$$
$$\rho $$ (r) $$\alpha $$ $${1 \over {{r^2}}}$$
Explanation
_10th_April_Morning_Slot_en_15_1.png)
Here, $$\Delta $$v and $$\Delta $$r are same for any pair of surface.
we know,
Electric field, E = $$-$$ $${{dv} \over {dr}}$$
$$ \therefore $$ E = constant [As dv and dr are constant]
Electric field inside the spherical charge distribution.
E = $${{\rho r} \over {3{\varepsilon _0}}}$$
Now, as E = constant
$$ \therefore $$ $$\rho $$ r = constant
$$ \Rightarrow $$ $$\rho $$ (r) $$ \propto $$ $${1 \over r}$$
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