JEE Main 2022 · Physics · Electrostatics · hard

A non-conducting solid sphere of radius R carries a volume charge density ρ=ρ0r/R at distance r from its centre. If E1 and E2 are the magnitudes of…

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A non-conducting solid sphere of radius

R

carries a volume charge density

\rho=\rho_0\dfrac{r}{R}

at distance

r

from its centre. If

E_1

and

E_2

are the magnitudes of the electric field at distances

\dfrac R2

and

\dfrac{3R}{2}

from the centre, then

\dfrac{E_1}{E_2}

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Answer: (C) $\dfrac{9}{16}$

Charge within radius

r\le R

q(r)=\int_0^r\rho_0\frac{r'}{R}4\pi r'^2dr'=\dfrac{\pi\rho_0r^{4}}{R}

. By Gauss's law

E=\dfrac{q}{4\pi\varepsilon_0r^{2}}=\dfrac{\rho_0r^{2}}{4\varepsilon_0R}

, so

E_1=\dfrac{\rho_0R}{16\varepsilon_0}

. Outside, the total charge is

\pi\rho_0R^{3}

, so

E=\dfrac{\rho_0R^{3}}{4\varepsilon_0r^{2}}

and

E_2=\dfrac{\rho_0R^{3}}{4\varepsilon_0}\cdot\dfrac{4}{9R^{2}}=\dfrac{\rho_0R}{9\varepsilon_0}

. Ratio

=\dfrac{1/16}{1/9}=\dfrac9{16}

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