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A spherical conducting shall of radius \mathrm{a}, centered at the origin, has a potential field \mathrm{V=\mathrm{ \begin{cases}V_0 & r \leq a \\ V_0 a / \delta & r<a\end{cases}}} with the zero reference at infinity. The stored energy potential to -

Option: 1

\mathrm{\frac{1}{2}\ QV}


Option: 2

\mathrm{QV}


Option: 3

\mathrm{QV\ ln\ 2}


Option: 4

None of the above / All are incorrect. 


Answers (1)

best_answer

\mathrm{w_E=\frac{1}{2} \int \epsilon_0\ E^2 d v \text {, where } E-\nabla v=\left\{\begin{array}{cc} 0 & r<a \\ \frac{v_0 a}{r^2}\ \bar{a}_r, & r>0 \end{array}\right.}
\mathrm{=\frac{\epsilon_0}{2} \int_0^{2 \pi} \int_0^\pi \int_0^{\infty}\left(\frac{V_0 Q}{r^2}\right)^2 r^2 \sin \theta\ d r\ d \theta\ d \phi}
\mathrm{=2 \pi\ t_0\ v_0^2\ a}
\mathrm{total\ charge\ Q=\Delta A=\left(\frac{\epsilon _0 v_0 a}{a^2}\right) \cdot 4 \pi a^2=4 \pi\ \epsilon_0\ v_0\ a}
\mathrm{ then,\ W_E=\frac{1}{2}\ Q V}

Posted by

himanshu.meshram

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