Electrostatics

JEE Main level, single correct, medium

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Question 122 of 311

[12]
A solid ball of radius R\mathrm{R} has a charge density ρ\rho given by ρ=ρ0(1−rR)\rho=\rho_0\left(1-\frac{r}{R}\right) for 0≤r≤R0 \leq r \leq R. The electric field outside the ball is:

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Answer: (D) ρ0R312ε0r2\frac{\rho_0 R^3}{12 \varepsilon_0 r^2}

Charge density, ρ=ρ0(1−rR)\rho=\rho_0\left(1-\frac{r}{R}\right)
dq=ρdvqin =∫dq=ρdv=ρ0(1−rR)4πr2dr(∵dv=4πr2dr)=4πρ0∫0R(1−rR)r2dr=4πρ0∫0Rr2dr−r2Rdr=4πρ0[[r33]0R−[r44R]0R]=4πρ0[R33−R44R]\begin{aligned} &d q=\rho d v \\ &q_{\text {in }}=\int d q=\rho d v \\ &=\rho_0\left(1-\frac{r}{R}\right) 4 \pi r^2 d r \quad\left(\because d v=4 \pi \mathrm{r}^2 \mathrm{dr}\right) \\ &=4 \pi \rho_0 \int_0^R\left(1-\frac{r}{R}\right) r^2 d r \\ &=4 \pi \rho_0 \int_0^R r^2 d r-\frac{r^2}{R} d r \\ &=4 \pi \rho_0\left[\left[\frac{r^3}{3}\right]_0^R-\left[\frac{r^4}{4 R}\right]_0^R\right] \\ &=4 \pi \rho_0\left[\frac{R^3}{3}-\frac{R^4}{4 R}\right] \end{aligned}
=4πρ0[R33−R34]=4πρ0[R312]=4 \pi \rho_0\left[\frac{R^3}{3}-\frac{R^3}{4}\right]=4 \pi \rho_0\left[\frac{R^3}{12}\right]
q=πρ0R33q=\frac{\pi \rho_0 R^3}{3}

E.4πr2=(πρ0R33ϵ0)E .4 \pi r^2=\left(\frac{\pi \rho_0 R^3}{3 \epsilon_0}\right)
∴\therefore \quad Electric field outside the ball, E=ρ0R312∈0r2E=\frac{\rho_0 R^3}{12 \in_0 r^2}

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