Question

# A solid nonconducting sphere of radius $$R$$ has a uniform charge distribution of volume charge density, $${ \rho }={ \rho }_{ 0 }\cfrac { r }{ R }$$, where $${ \rho }_{ 0 }$$ is a constant and r is the distance from the centre of the sphere. Show that:(a) the total charge on the sphere is $$Q=\pi { \rho }_{ 0 }{ R }^{ 3 }$$(b) the electric field inside the sphere has a magnitude given by,$$E=\cfrac { KQ{ r }^{ 2 } }{ { R }^{ 4 } }$$

Solution

## a) : Total charge , $$Q=\int_0^R\rho dV=\int^R_0 \frac{\rho_0 r}{R}4\pi r^2 dr=\dfrac{4\pi\rho_0}{R}[R^4/4]=\pi \rho_0R^3$$b) By Gauss's law, the field inside sphere, $$E.4\pi r^2=\dfrac{Q_{en}}{\epsilon_0}$$Here, $$Q_{en}=\int_0^r\rho dV=\int^r_0 \frac{\rho_0 r}{R}4\pi r^2 dr=\dfrac{4\pi\rho_0}{R}[r^4/4]=\pi \rho_0r^4/R=Qr^4/R^4$$So,  $$E.4\pi r^2=\dfrac{Qr^4}{R^4\epsilon_0}$$or $$E=\dfrac{KQr^2}{R^4}$$ where $$K=\dfrac{1}{4\pi \epsilon_0}$$Physics

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