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Step-by-Step Solution
Step 1: Equate the surface charge densities of the spheres
The smaller sphere has radius $R$ and carries charge $Q_1$, so its surface area is $4\pi R^2$ and its surface charge density is
$\sigma_1 = \frac{Q_1}{4\pi R^2}.$
The larger sphere has radius $4R$ and carries charge $Q_2$, so its surface area is $4\pi\,(4R)^2 = 64\pi R^2,$ and its surface charge density is
$\sigma_2 = \frac{Q_2}{64\pi R^2}.$
Given that $\sigma_1 = \sigma_2$, we have:
$\frac{Q_1}{4\pi R^2} \;=\; \frac{Q_2}{64\pi R^2}
\quad\Longrightarrow\quad
Q_2 \;=\; 16\,Q_1.$
Step 2: Write expressions for potentials at radii R and 4R
Since these are conducting spheres, the potential at the surface of a conductor is given by
$V = \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{r},$
for that sphere and also includes contributions from other charges.
Potential at $r = R$ (the smaller sphere):
This potential is the sum of the potential due to its own charge $Q_1$ plus the potential due to the outer sphere’s charge $Q_2$. The potential inside (and on the surface of) a spherical shell of radius $4R$ carrying charge $Q_2$ is the same as on its surface. Hence:
$V(R)
= \frac{1}{4\pi\varepsilon_0}\,\frac{Q_1}{R}
\;+\;
\frac{1}{4\pi\varepsilon_0}\,\frac{Q_2}{4R}
\;=\;
\frac{Q_1}{4\pi\varepsilon_0\,R}
\;+\;
\frac{Q_2}{16\pi\varepsilon_0\,R}.
$
Potential at $r = 4R$ (the larger sphere):
Here, it is the sum of the potential due to $Q_1$ at distance $4R$ plus the potential of the outer sphere due to its own charge $Q_2$ on its surface:
$V(4R)
= \frac{1}{4\pi\varepsilon_0}\,\frac{Q_1}{4R}
\;+\;
\frac{1}{4\pi\varepsilon_0}\,\frac{Q_2}{4R}
=
\frac{Q_1 + Q_2}{4\pi\varepsilon_0\,(4R)}.
$
Step 3: Substitute $Q_2 = 16\,Q_1$ into the expression
Before substituting, compute the difference:
$\displaystyle
V(R) - V(4R)
\,=\,
\biggl(\frac{Q_1}{4\pi\varepsilon_0\,R} + \frac{Q_2}{16\pi\varepsilon_0\,R}\biggr)
\;-\;
\frac{Q_1 + Q_2}{16\pi\varepsilon_0\,R}.
$
Carefully simplifying,
$V(R) - V(4R)
= \frac{Q_1}{4\pi\varepsilon_0\,R}
\;+\;
\frac{Q_2}{16\pi\varepsilon_0\,R}
\;-\;
\frac{Q_1 + Q_2}{16\pi\varepsilon_0\,R}.
$
Rewriting, we see the $Q_2$ terms cancel out:
$
= \frac{Q_1}{4\pi\varepsilon_0\,R}
-
\frac{Q_1}{16\pi\varepsilon_0\,R}
= \frac{Q_1}{4\pi\varepsilon_0\,R}\,\Bigl(1 - \tfrac{1}{4}\Bigr)
= \frac{Q_1}{4\pi\varepsilon_0\,R}\,\cdot\,\frac{3}{4}
= \frac{3\,Q_1}{16\pi\varepsilon_0\,R}.
$
Step 4: Conclude the final answer
Therefore, the potential difference
$V(R) - V(4R)$
is
$\displaystyle \frac{3Q_1}{16\pi \varepsilon_0\,R}.$
