Engineering
Physics
Electric Flux and Gauss Law and its Applications
Question
The electrostatic potential inside a charged spherical ball is given by V= ar2 + b where r is the distance from the centre ; a, b are constant. Then the charge density inside the ball is :
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Solution
\(\oint {\vec E.\overrightarrow {ds} } = \frac{q}{{{\varepsilon _0}}}\)
\( - 2ar.4a{r^2} = \frac{q}{{{\varepsilon _0}}}\)
q = –8 e0apr3
\(\rho = \frac{q}{{\frac{4}{3}\pi {r^3}}}\)
r = –6ae0