Three concentric spherical shells have radii a, \(b\) and \(c(a

Short Answer

Expert verified
(D) \(\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}}\)

Step by step solution

01

Write the expression for potential

For a spherical shell with a radius r and surface charge density σ, the potential V can be written as: \[V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}\] where Q is the total charge on the shell, and \(\epsilon_0\) is the vacuum permittivity. For our problem, we will denote the total charge for each shell as \(Q_A\), \(Q_B\), and \(Q_C\).
02

Find the total charge on each shell

The total charge on a spherical shell is given by: \[Q = \sigma \times \text{Surface Area}\] For each shell, their total charges are: - \(Q_A = \sigma \times 4\pi a^2\) - \(Q_B = -\sigma \times 4\pi b^2\) - \(Q_C = \sigma \times 4\pi (a+b)^2\)
03

Calculate the potentials for each shell

Now, we can substitute the corresponding total charges into the expression for potential to calculate the potentials for each shell: - \(V_A = \frac{1}{4 \pi \epsilon_0} \frac{\sigma \times 4\pi a^2}{a} = \frac{\sigma a}{\epsilon_0}\) - \(V_B = \frac{1}{4 \pi \epsilon_0} \frac{-\sigma \times 4\pi b^2}{b} = -\frac{\sigma b}{\epsilon_0}\) - \(V_C = \frac{1}{4 \pi \epsilon_0} \frac{\sigma \times 4\pi (a+b)^2}{(a + b)} = \frac{\sigma (a+b)}{\epsilon_0}\)
04

Compare the potentials

Now, we can compare the potentials of the shells: - \(V_C = \frac{\sigma (a+b)}{\epsilon_0} = \frac{\sigma a}{\epsilon_0} + \frac{\sigma b}{\epsilon_0} = V_A - V_B\) From this result, we can see that \(V_C = V_A \neq V_B\), which corresponds to option (D). Therefore, the correct answer is: (D) \(\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}}\)

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