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Step-by-Step Solution
Step 1: Understand the System of Charges
We have three concentric spherical shells of radii $a$, $b$, and $c$, with $a < b < c$. Their surface charge densities are:
Shell A (radius $a$): $+\sigma$
Shell B (radius $b$): $-\sigma$
Shell C (radius $c$): $+\sigma$
Hence, the total charges on the shells are:
$$Q_A = 4\pi a^2 \sigma,\quad Q_B = -\,4\pi b^2 \sigma,\quad Q_C = 4\pi c^2 \sigma.$$
Step 2: Formula for Potential on Each Shell
For a spherical shell, the potential on the shell due to:
Its own charge $Q$ (on radius $R$) is $ \frac{1}{4\pi \epsilon_0}\, \frac{Q}{R}.$
Another concentric shell of radius $R'$ (where $R' > R$) is $ \frac{1}{4\pi \epsilon_0}\, \frac{Q'}{R'} $ if the shell lies outside the point where we measure the potential.
Another concentric shell of radius $R''$ (where $R'' < R$) is $ \frac{1}{4\pi \epsilon_0}\, \frac{Q''}{R} $ if the shell lies inside.
Using these principles, we can directly write the potentials on each shell.
Step 3: Potential on Shell A ($V_A$)
Shell A (radius $a$) is inside shells B and C. Therefore:
$$
V_A \;=\; \frac{1}{4\pi \epsilon_0} \left( \frac{Q_A}{a} + \frac{Q_B}{b} + \frac{Q_C}{c} \right).
$$
Substitute $Q_A = 4\pi a^2 \sigma$, $Q_B = -4\pi b^2 \sigma$, $Q_C = 4\pi c^2 \sigma$:
$$
V_A = \frac{1}{4\pi \epsilon_0} \Biggl(\frac{4\pi a^2 \sigma}{a} \;+\; \frac{-\,4\pi b^2 \sigma}{b} \;+\; \frac{4\pi c^2 \sigma}{c}\Biggr).
$$
Factor out $\frac{4\pi \sigma}{4\pi \epsilon_0}$:
$$
V_A = \frac{1}{\epsilon_0}\,\bigl(a\,\sigma - b\,\sigma + c\,\sigma \bigr).
$$
Step 4: Potential on Shell B ($V_B$)
Shell B (radius $b$) is outside shell A but inside shell C. So the potential on B consists of:
Potential due to $Q_A$ at radius $b$.
Potential of its own charge $Q_B$ at radius $b$.
Potential due to $Q_C$ at radius $c$.
Hence,
$$
V_B = \frac{1}{4\pi \epsilon_0}\left(\frac{Q_A}{b} + \frac{Q_B}{b} + \frac{Q_C}{c}\right).
$$
Substitute the charges:
$$
V_B = \frac{1}{4\pi \epsilon_0}\left(\frac{4\pi a^2 \sigma}{b} + \frac{-4\pi b^2 \sigma}{b} + \frac{4\pi c^2 \sigma}{c}\right),
$$
which simplifies to:
$$
V_B = \frac{1}{\epsilon_0}\left(\frac{a^2 \sigma}{b} - b\,\sigma + c\,\sigma \right).
$$
Step 5: Potential on Shell C ($V_C$)
Shell C (radius $c$) encloses both shells A and B. The potential on C is:
$$
V_C = \frac{1}{4\pi \epsilon_0}\left(\frac{Q_A}{c} + \frac{Q_B}{c} + \frac{Q_C}{c}\right).
$$
Substitute the charges:
$$
V_C = \frac{1}{4\pi \epsilon_0}\Bigl(\frac{4\pi a^2 \sigma}{c} + \frac{-4\pi b^2 \sigma}{c} + \frac{4\pi c^2 \sigma}{c}\Bigr),
$$
which simplifies to:
$$
V_C = \frac{1}{\epsilon_0}\,\bigl(\frac{a^2 \sigma}{c} - \frac{b^2 \sigma}{c} + c\,\sigma\bigr).
$$
Step 6: Apply the Condition $c = a + b$ and Compare
We are given $c = a + b$. One way to see the equality of $V_A$ and $V_C$ explicitly is to substitute $c = a + b$ into the expressions for $V_A$ and $V_C$ and simplify. When worked through (either by factoring or plugging in example numeric ratios that respect $c = a + b$), it turns out:
$$
V_A = V_C \;\;\text{and}\;\; V_B \text{ has a different value}.
$$
Thus, the potentials on the innermost and outermost shells are equal, but the middle shell's potential differs.
Final Conclusion
Therefore, the relationship among the potentials is:
$$
V_C = V_A \;\ne\; V_B.
$$
This matches the provided correct answer.
