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When calculating the electric potential inside the hollow portion why are we adding all the electric potentials of the previous cases??? Instead when calculating for other cases i.e the cases before this hollow portion case we only consider the electric potential of that case only.

See the image for more clearer meaning..... Also please note that only one charge +q is present inside the conducting hollow sphere.enter image description here

Qmechanic
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Tathagata Majumdar
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    Could you clarify exactly what you are confused about? – fhorrobin Jun 22 '18 at 18:26
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    Please don't post questions that depend on a cell phone image. This creates problems for visually impaired users and defeats the search features of the site. –  Jun 22 '18 at 20:14
  • To reiterate, please edit the question to remove the text from the images. We want search engines to be able to index the content of our questions. Furthermore, text is easier to see on displays of all sizes. In its present state, this question may be closed as Very Low Quality due to formatting issues. You can't reasonably expect somebody else to type out the content, so we'd like it if you could condense all that content from the book into a few useful sentences. Feel free to include images as necessary, but they cannot contain large blocks of text and math. –  Jun 24 '18 at 09:16

2 Answers2

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The space inside a hollow sphere of radius $\:R\:$ with charge $\:Q\:$ uniformly distributed on its surface [surface charge density $\:\sigma=Q/(4\pi R^2\:$)] is an equipotential region. That is with your choice of $\:V=0\:$ as $\:r\longrightarrow \infty\:$ the potential is $\:V_{inside}=0\:$. The potential outside is that of a point charge $\:Q\:$ positioned at the center of the sphere \begin{equation} V= \begin{cases} \hphantom{\dfrac{1}{4\pi}}0 & \text{for}\quad r<R\\ \dfrac{1}{4\pi \epsilon_{0}}\dfrac{Q}{r} & \text{for}\quad r\ge R \end{cases} \tag{01} \end{equation}

Related : see my answer as user82794 (former diracpaul) here Would you be weightless at the center of the Earth?

Frobenius
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Outside the shell, at a distance $r>b$, the potential $V(r)$ would be: $$ V(r)= \frac{kq}{r} + \frac{k(-q)}{r} + \frac{kq}{r} $$ The first term in the above expression corresponds to the potential due to point charge, the second corresponds to potential due to charges induced on the inner surface of the conductor and the third term corresponds to the potential due to the charges induced on the outer surface of the conductor, all at a distance $r \geq b$ from the centre. As we see that the potential due to point charge and charges induced on the inner surface are equal and opposite, we need not mention them separately in the expression for $V(r)$ as they cancel out.

That’s what the solution intends to convey for potential $V(r)$ for $a \leq r < b$ and $r<a$ too.

P.S. $k= \frac{1}{4πε₀}$

Aura Sartori
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