- JEE Mains 2014
- Hard
- MCQ
A spherically symmetric charge distribution is characterised by a charge density having the following variations \(\rho (r)\, = \,{\rho _0}\left( {1 - \frac{r}{R}} \right)\) for \(r < R\) \(\rho (r)\,=\,0\) for \(r\, \ge \,R\) Where \(r\) is the distance from the centre of the charge distribution \(\rho _0\) is a constant. The electric field at an internal point \((r < R)\) is
Answer & Solution
(B) \(\frac{{{\rho _0}}}{{{\varepsilon _0}}}\left( {\frac{r}{3} - \frac{{{r^2}}}{{4R}}} \right)\)
Step-by-step Solution
Detailed explanation
Charge on this shell
\(\mathrm{dq}=\rho 4 \pi \mathrm{x}^{2} \mathrm{dx}=\rho_{0}\left(1-\frac{\mathrm{x}}{\mathrm{R}}\right) .4 \pi \mathrm{x}^{2} \mathrm{dx}\)
\(\therefore\) Total charge in the spherical region from centre to \(r(r < R)\)
\(q=\int d q=4 \pi \rho_{0} \int_{0}^{T}\left(1-\frac{x}{R}\right) x^{2} d x\)
\(=4 \pi \rho_{0}\left[\frac{x^{3}}{3}-\frac{x^{4}}{4 R}\right]_{0}^{t}\)
\(=4 \pi \rho_{0}\left[\frac{r^{3}}{3}-\frac{r^{4}}{4 R}\right]\)
\(=4 \pi \rho_{0} r^{3}\left[\frac{1}{3}-\frac{r}{4 R}\right]\)
\(\therefore\) Electric field atr, \(\mathrm{E}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{r}^{2}}\)
\( = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{4\pi {\rho _0}{r^3}}}{{{r^2}}} \cdot \left[ {\frac{1}{3} - \frac{r}{{4R}}} \right]\)
\(=\frac{\rho_{0}}{\varepsilon_{0}}\left[\frac{\tau}{3}-\frac{\tau^{2}}{4 R}\right]\)
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