A spherically symmetric charge distribution is considered with charge density varying as (r)= array ll 0 (3/4-r/R) & for r R Zero & for r>R array. 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)

(D) \frac{\rho_{0} r}{5 \varepsilon_{0}}\left(1-\frac{r}{R}\right)

JEE Mains · Physics · STD 12 - 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.

Pick an option to check your answer.

Verified Solution

Correct Answer

(C) \(\frac{\rho_{0} r}{4 \varepsilon_{0}}\left(1-\frac{r}{R}\right)\)

Detailed explanation

\(\oint \overrightarrow{ E } \cdot d \overrightarrow{ s }=\frac{ Q _{\text {in }}}{\varepsilon_{ o }}\)

\(E .4 \pi r ^{2}=\frac{\int_{0}^{ r } \rho_{ o }\left(\frac{3}{4}-\frac{ r }{ R }\right) 4 \pi r ^{2} dr }{\varepsilon_{0}}\)

\(E 4 \pi r ^{2}=\frac{\rho_{ o } 4 \pi}{\varepsilon_{ o }}\left(\frac{3}{4} \frac{ r ^{3}}{3}-\frac{ r ^{4}}{4 R }\right)\)

\(Er { }^{2}=\frac{\rho_{ o } r ^{3}}{4 \varepsilon_{ o }}\left\{1-\frac{ r }{ R }\right\}\)

\(E =\frac{\rho_{0} r }{4 \varepsilon_{ o }}\left\{1-\frac{ r }{ R }\right\}\)

Same subject

List\(-I\)	List\(-II\)
\((a)\) \(\omega L\,>\,\frac{1}{\omega C}\)	\((i)\) Current is in phase with \(emf\)
\((b)\) \(\omega {L}=\frac{1}{\omega {C}}\)	\((ii)\) Current lags behind the applied \(emf\)
\((c)\) \(\omega {L}\, < \,\frac{1}{\omega {C}}\)	\((iii)\) Maximum current occurs
\((d)\) resonant frequency	\((iv)\) Current leads the \(emf\)

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