Electrostatics

JEE Main level, single correct, medium

137 of 311

Question 137 of 311

Let there be a spherically symmetric charge distribution with charge density varying as ρ(r)=ρ0(54−rR)\rho(r)=\rho_0\left(\frac{5}{4}-\frac{r}{R}\right) upto r=Rr=R, and ρ(r)=0\rho(r)=0 for r>Rr\gt R, where rr is the distance from the origin. The electric field at a distance r(r<R)r(r \lt R) from the origin is given by

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Answer: (B) ρ0r4ε0(53−rR)\frac{\rho_0 r}{4 \varepsilon_0}\left(\frac{5}{3}-\frac{r}{R}\right)

Apply shell theorem the total charge upto distance rr can be calculated as followed dq=4πr2⋅dr.ρd q=4 \pi r^2 \cdot d r . \rho
=4πr2⋅dr⋅ρ0[54−rR]=4πρ0[54r2dr−r3Rdr]∫dq=q=4πρ0∫0r(54r2dr−r3Rdr)=4πρ0[54r33−1Rr44]E=kqr2=14πε01r2⋅4πρ0[54(r33)−r44R]E=ρ0r4ε0[53−rR]\begin{aligned} & =4 \pi r^2 \cdot d r \cdot \rho_0\left[\frac{5}{4}-\frac{r}{R}\right] \\ & =4 \pi \rho_0\left[\frac{5}{4} r^2 d r-\frac{r^3}{R} d r\right] \\ & \int d q=q=4 \pi \rho_0 \int_0^r\left(\frac{5}{4} r^2 d r-\frac{r^3}{R} d r\right) \\ & =4 \pi \rho_0\left[\frac{5}{4} \frac{r^3}{3}-\frac{1}{R} \frac{r^4}{4}\right] \\ & E=\frac{k q}{r^2} \\ & =\frac{1}{4 \pi \varepsilon_0} \frac{1}{r^2} \cdot 4 \pi \rho_0\left[\frac{5}{4}\left(\frac{r^3}{3}\right)-\frac{r^4}{4 R}\right] \\ & E=\frac{\rho_0 r}{4 \varepsilon_0}\left[\frac{5}{3}-\frac{r}{R}\right] \\ & \end{aligned}

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