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Step-by-Step Solution
Step 1: Identify the Charges and the Region of Interest
We have two charges:
1. A charge $Q$ placed at the center of a thin spherical conducting shell of radius $R$.
2. A charge $q$ distributed (effectively) on the surface of the spherical shell.
We want the potential at a point $P$ located at a distance $R/2$ from the center. Because $R/2 < R$, point $P$ lies inside the spherical shell.
Step 2: Potential Due to Charge $Q$ at the Center
For a point charge $Q$ located at the center of the sphere, the potential at any point inside is calculated just like the potential in free space:
$V_1 = \dfrac{1}{4\pi \varepsilon_0}\,\dfrac{Q}{\text{distance from charge}}.$
Since $P$ is at distance $R/2$ from $Q$,
$V_1 = \dfrac{1}{4\pi \varepsilon_0}\,\dfrac{Q}{R/2} = \dfrac{2Q}{4\pi \varepsilon_0\,R}.$
Step 3: Potential Due to Charge $q$ on the Spherical Shell
For a conducting shell, all excess charge $q$ resides on its surface. The potential inside a conductor due to charges on its surface is the same everywhere inside, and it is given by:
$V_2 = \dfrac{1}{4\pi \varepsilon_0}\,\dfrac{q}{R}.$
This result holds true for every point inside the spherical shell.
Step 4: Total Potential at Point $P$
The superposition principle states that the total potential is the sum of the potentials due to each charge taken separately. Hence,
$V_{\text{total}} = V_1 + V_2.$
Substituting the expressions from above:
$V_{\text{total}}
= \dfrac{2Q}{4\pi \varepsilon_0\,R}
\;+\; \dfrac{q}{4\pi \varepsilon_0\,R}
= \dfrac{2Q + q}{4\pi \varepsilon_0\,R}.$
Step 5: Final Answer
Therefore, the electrostatic potential at the point $P$ a distance $R/2$ from the center is
$ \displaystyle \dfrac{2Q}{4\pi \varepsilon_0 \, R} + \dfrac{q}{4\pi \varepsilon_0 \, R}.$
