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  • Let P(r)=\frac{Q}{\pi R^{4}}r be the charge density distribution for a solid sphere of radius R and total charge Q . For a point 'p' insid

Let P(r)=\frac{Q}{\pi R^{4}}r be the charge density distribution for a solid sphere of radius R and total charge Q . For a point 'p' inside the sphere at a distance r1 from the centre of the sphere.

Answers (1)

Given that, The charge density \rho(r)=\frac{Q r}{4 \pi R^{4}} 
According to the fig,
The electric field at the point P distant r_{1}  from the center,
According to gauss's theorem $E \times 4 \pi r_{1}^{2}=$ \text{charge enclosed}/\varepsilon $_{0}$

\begin{array}{l} E \times 4 \pi r_{1}^{2}=\frac{1}{\varepsilon_{0}} \int \rho d V \\ E \times 4 \pi r_{1}^{2}=\frac{1}{\varepsilon_{0}} \int_{0}^{r_{1}} \frac{Q r}{4 \pi R^{4}} \times 4 \pi r^{2} d r \\ E=\frac{1}{\varepsilon_{0}} \times \frac{Q r_{1}^{4}}{4 \times 4 \pi R^{4}} \times 4 \pi \times \frac{1}{4 \pi r_{1}^{2}} \\ E=\frac{Q r_{1}^{2}}{\varepsilon_{0} 16 \pi R^{4}} \end{array}

Therefore, the electric field at a point $P$ will be ; E=\frac{Q r_{1}^{2}}{\varepsilon_{0} 16 \pi R^{4}}.

Posted by

Satyajeet Kumar

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