JEE Main 2020 · Physics · Electrostatics · hard

A non-conducting sphere of radius R carries a volume charge density ρ = ρ0 r/R (where r is the distance from the centre) for r ≤ R, and there is no…

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A non-conducting sphere of radius

R

carries a volume charge density

\rho = \rho_0\, r/R

(where

r

is the distance from the centre) for

r \le R

, and there is no other charge. If

E_1

is the electric field magnitude at

r = R/2

and

E_2

the magnitude at

r = 2R

, then

E_1/E_2

equals

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Answer: (A) $1$

Enclosed charge inside radius

r \le R

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

. Gauss's law:

E(r) = \dfrac{q}{4\pi\varepsilon_0 r^2} = \dfrac{\rho_0 r^2}{4\varepsilon_0 R}

, so

E_1 = \dfrac{\rho_0 R}{16\varepsilon_0}

. Total charge

Q = \pi\rho_0R^3

, so

E_2 = \dfrac{Q}{4\pi\varepsilon_0 (2R)^2} = \dfrac{\rho_0 R}{16\varepsilon_0}

. The ratio is

1

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