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← JEE Main 2022 · 29 Jul · Shift 1
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JEE Main 2022 · 29 Jul · Shift 1, Physics question 40

A spherically symmetric charge distribution is considered with charge density varying as

ρ(r)={ρ0(34−rR)for r≤Rzerofor r>R\rho(r)=\begin{cases}\rho_0\left(\dfrac34-\dfrac rR\right) & \text{for } r\le R\\ \text{zero} & \text{for } r>R\end{cases}

Where, r (r < R) is the distance from the centre O (as shown in figure). The electric field at point P will be :

A sphere of radius R with charge density ρ(r); a point P inside at distance r from the centre O.
Choose one option
Answer and worked solution

Official answer: C

Enclosed charge ∫0rρ0(34−sR)4πs2ds=πρ0r3(1−rR)\displaystyle\int_0^r\rho_0\left(\frac34-\frac sR\right)4\pi s^2ds=\pi\rho_0r^3\left(1-\frac rR\right).

Gauss E=Q4πε0r2=ρ0r4ε0(1−rR)E=\dfrac{Q}{4\pi\varepsilon_0r^2}=\dfrac{\rho_0r}{4\varepsilon_0}\left(1-\dfrac rR\right).

(C)

Topics:Electric flux and Gauss's law
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