A) \[-6\,\,a{{\varepsilon }_{0}}\]
B) \[-24\,\pi \,a{{\varepsilon }_{0}}\gamma \]
C) \[-6\,a{{\varepsilon }_{0}}\gamma \]
D) \[-24\,\,\pi \,a{{\varepsilon }_{0}}\gamma \]
Correct Answer: A
Solution :
\[\phi =a{{r}^{2}}+b\] \[\frac{d\phi }{dr}=-2ar\Rightarrow E=-2ar\] \[E\times 4\pi {{r}^{2}}=\frac{q}{{{\varepsilon }_{0}}}\] \[\Rightarrow \,\,q=8\pi {{\varepsilon }_{0}}a{{r}^{3}}\] \[\Rightarrow \,\,\rho =\frac{dq}{dV}\] \[=\frac{-24\pi {{\varepsilon }_{0}}{{r}^{2}}dr}{4\pi {{r}^{2}}dr}\] \[=6{{\varepsilon }_{0}}a\]
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