Three concentric spherical shells have radii `a, b` and `c(a lt b lt c)` and have surface charge densities `sigma, -sigam` and `sigma` respectively. If `V_(A), V_(B)` and `V_(C)` denote the potentials of the three shells, then for `c = a + b`, we have

Options

A. `V_(C)=V_(B)=V_(A)`
B. `V_(C)=V_(A)ne V_(B)`
C. `V_(C)= V_(B)ne V_(A)`
D. `V_(C) ne V_(B) ne V_(A)`

Answer

To solve the problem, we need to find the potentials \( V_A \), \( V_B \), and \( V_C \) for three concentric spherical shells with given surface charge densities and radii. The shells have radii \( a \), \( b \), and \( c \) (where \( a < b < c \)) and surface charge densities \( \sigma \), \( -\sigma \), and \( \sigma \) respectively. We are also given that \( c = a + b \).

### Step 1: Calculate the potential \( V_A \) at radius \( a \)

The potential \( V_A \) due to the three shells can be calculated using the formula for the potential due to a charged shell:

\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{R} \]

Where \( Q \) is the total charge on the shell and \( R \) is the distance from the center.

1. The charge on the first shell (radius \( a \)) is:
\[ Q_A = \sigma \cdot 4\pi a^2 \]
Therefore, the potential \( V_A \) due to this shell is:
\[ V_A = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_A}{a} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{\sigma \cdot 4\pi a^2}{a} = \frac{\sigma a}{\epsilon_0} \]

2. The potential due to the second shell (radius \( b \)) is negative because it has a charge density of \( -\sigma \):
\[ Q_B = -\sigma \cdot 4\pi b^2 \]
Thus, the potential \( V_B \) is:
\[ V_B = \frac{1}{4 \pi \epsilon_0} \cdot \frac{-\sigma \cdot 4\pi b^2}{a} = -\frac{\sigma b^2}{\epsilon_0 a} \]

3. The potential due to the third shell (radius \( c \)) is:
\[ Q_C = \sigma \cdot 4\pi c^2 \]
Therefore, the potential \( V_C \) is:
\[ V_C = \frac{1}{4 \pi \epsilon_0} \cdot \frac{\sigma \cdot 4\pi c^2}{a} = \frac{\sigma c^2}{\epsilon_0 a} \]

Combining these, we find \( V_A \):
\[ V_A = \frac{\sigma a}{\epsilon_0} - \frac{\sigma b^2}{\epsilon_0 a} + \frac{\sigma c^2}{\epsilon_0 a} \]

### Step 2: Calculate the potential \( V_B \) at radius \( b \)

Using the same approach, we calculate \( V_B \):
1. The potential due to the first shell (radius \( a \)):
\[ V_A = \frac{\sigma a}{\epsilon_0 b} \]

2. The potential due to the second shell (radius \( b \)):
\[ V_B = -\frac{\sigma b^2}{\epsilon_0 b} = -\frac{\sigma b}{\epsilon_0} \]

3. The potential due to the third shell (radius \( c \)):
\[ V_C = \frac{\sigma c^2}{\epsilon_0 b} \]

Combining these, we find \( V_B \):
\[ V_B = \frac{\sigma a}{\epsilon_0 b} - \frac{\sigma b}{\epsilon_0} + \frac{\sigma c^2}{\epsilon_0 b} \]

### Step 3: Calculate the potential \( V_C \) at radius \( c \)

Similarly, for \( V_C \):
1. The potential due to the first shell (radius \( a \)):
\[ V_A = \frac{\sigma a}{\epsilon_0 c} \]

2. The potential due to the second shell (radius \( b \)):
\[ V_B = -\frac{\sigma b^2}{\epsilon_0 c} \]

3. The potential due to the third shell (radius \( c \)):
\[ V_C = \frac{\sigma c^2}{\epsilon_0 c} = \frac{\sigma c}{\epsilon_0} \]

Combining these, we find \( V_C \):
\[ V_C = \frac{\sigma a}{\epsilon_0 c} - \frac{\sigma b^2}{\epsilon_0 c} + \frac{\sigma c}{\epsilon_0} \]

### Step 4: Analyze the results

Now we can analyze the results:
- We have \( V_A \), \( V_B \), and \( V_C \).
- Since \( c = a + b \), we can substitute this into our equations to find relationships between \( V_A \), \( V_B \), and \( V_C \).

### Final Result

After simplification, we find that:
\[ V_A = V_C \quad \text{and} \quad V_B \neq V_A, V_C \]

Thus, the final conclusion is:
\[ V_A = V_C \quad \text{and} \quad V_B \neq V_A = V_C \]