JEE Mains · Physics · STD 12 - 1. Electric charges and fields
A spherically symmetric charge distribution is considered with charge density varying as \(\rho(r)=\left\{\begin{array}{ll}\rho_{0}\left(\frac{3}{4}-\frac{r}{R}\right) & \text { for } r \leq R \\ \text { Zero } & \text { for } r>R\end{array}\right.\) Where, \(r ( r < R )\) is the distance from the centre \(O\) (as shown in figure). The electric field at point \(P\) will be.

- A \(\frac{\rho_{0} r}{4 \varepsilon_{0}}\left(\frac{3}{4}-\frac{r}{R}\right)\)
- B \(\frac{\rho_{0} r}{3 \varepsilon_{0}}\left(\frac{3}{4}-\frac{r}{R}\right)\)
- C \(\frac{\rho_{0} r}{4 \varepsilon_{0}}\left(1-\frac{r}{R}\right)\)
- D \(\frac{\rho_{0} r}{5 \varepsilon_{0}}\left(1-\frac{r}{R}\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{\rho_{0} r}{4 \varepsilon_{0}}\left(1-\frac{r}{R}\right)\)
Step-by-step Solution
Detailed explanation
\(\oint \overrightarrow{ E } \cdot d \overrightarrow{ s }=\frac{ Q _{\text {in }}}{\varepsilon_{ o }}\) \(E .4 \pi r ^{2}=\frac{\int_{0}^{ r } \rho_{ o }\left(\frac{3}{4}-\frac{ r }{ R }\right) 4 \pi r ^{2} dr }{\varepsilon_{0}}\)…
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