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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2016
  • Hard
  • MCQ

The region between two concentric spheres ofradii '\(a\)' and '\(b\)', respectively (see figure), have volume charge density \(\rho = \frac{A}{r}\) where \(A\) is a constant and \(r\) is the distance from the centre. At the centre of the spheres is a point charge \(Q\). The value of \(A\) such that the electric field in the region between the spheres will be constant, is :

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Answer & Solution

Correct Answer

(C) \(\;\frac{Q}{{2\pi {a^2}}}\)

Step-by-step Solution

Detailed explanation

Applying Gauss's law

\(\oint_{\mathrm{S}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}=\frac{Q}{\epsilon_{0}}\)

\(\therefore \mathrm{E} \times 4 \pi \mathrm{r}^{2}=\frac{\mathrm{Q}+4 \pi \mathrm{ar}^{2}-4 \pi \mathrm{Aa}^{2}}{\epsilon_{0}}\)

\(\rho=\frac{\mathrm{d} \mathrm{r}}{\mathrm{d} \mathrm{v}}\)

\(Q=\rho 4 \pi r^{2}\)

\(\mathrm{Q}=\int_{\mathrm{a}}^{\mathrm{A}} \frac{\mathrm{A}}{\mathrm{r}} 4 \pi \mathrm{r}^{2} \mathrm{dr}=4 \pi \mathrm{A}\left[\mathrm{r}^{2}-\mathrm{a}^{2}\right]\)

\(\mathrm{E}=\frac{1}{4 \pi \epsilon_{0}}\left[\frac{\mathrm{Q}-4 \pi \mathrm{Aa}^{2}}{\mathrm{r}^{2}}+4 \pi \mathrm{A}\right]\)

For \(\mathrm{E}\) to be independent of  \('{r}'\)

\(\mathrm{Q}-2 \pi \mathrm{Aa}^{2}=0\)

\(\therefore A=\frac{Q}{2 \pi a^{2}}\)

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