Two concentric conducting spherical shells of radii `a_1 and a_2 (a_2 gt a_1)` are charged to potentials `phi_1` and `phi_2`, respectively. Find the charge on the inner shell.

Options

A. `q_1 = 4 pi epsilon_0 ((phi_1 - phi_2)/(a_2 - a_1)) a_1 a_2`
B. `q_1 = 4 pi epsilon_0 ((phi_1 + phi_2)/(a_2 + a_1)) a_1 a_2`
C. `q_1 = pi epsilon_0 ((phi_1 - phi_2)/(a_2 + a_1)) a_1 a_2`
D. `q_1 = 2 pi epsilon_0 ((phi_1 + phi_2)/(a_2 - a_1)) a_1 a_2`

Explanation

To find the charge on the inner shell of two concentric conducting spherical shells with radii \( a_1 \) and \( a_2 \) (where \( a_2 > a_1 \)) charged to potentials \( \phi_1 \) and \( \phi_2 \) respectively, we can follow these steps:

### Step 1: Understand the Potential of Each Shell
The potential \( \phi_1 \) at the surface of the inner shell (radius \( a_1 \)) due to its own charge \( Q_1 \) and the charge \( Q_2 \) on the outer shell (radius \( a_2 \)) can be expressed as:
\[ \phi_1 = k \frac{Q_1}{a_1} + k \frac{Q_2}{a_2} \]
where \( k \) is the Coulomb's constant \( k = \frac{1}{4 \pi \epsilon_0} \).

### Step 2: Write the Expression for the Outer Shell
Similarly, the potential \( \phi_2 \) at the surface of the outer shell (radius \( a_2 \)) can be expressed as:
\[ \phi_2 = k \frac{Q_1}{a_2} + k \frac{Q_2}{a_2} \]

### Step 3: Set Up the Equation for Potential Difference
We can rearrange the equations for \( \phi_1 \) and \( \phi_2 \):
- From the equation for \( \phi_1 \):
\[ \phi_1 = k \frac{Q_1}{a_1} + k \frac{Q_2}{a_2} \]
- From the equation for \( \phi_2 \):
\[ \phi_2 = k \frac{Q_1}{a_2} + k \frac{Q_2}{a_2} \]

### Step 4: Subtract the Two Equations
Now, subtract the equation for \( \phi_2 \) from \( \phi_1 \):
\[ \phi_1 - \phi_2 = \left(k \frac{Q_1}{a_1} + k \frac{Q_2}{a_2}\right) - \left(k \frac{Q_1}{a_2} + k \frac{Q_2}{a_2}\right) \]
This simplifies to:
\[ \phi_1 - \phi_2 = k \left( \frac{Q_1}{a_1} - \frac{Q_1}{a_2} \right) \]

### Step 5: Factor Out \( Q_1 \)
Factor out \( Q_1 \):
\[ \phi_1 - \phi_2 = k Q_1 \left( \frac{1}{a_1} - \frac{1}{a_2} \right) \]

### Step 6: Solve for \( Q_1 \)
Rearranging gives us:
\[ Q_1 = \frac{(\phi_1 - \phi_2) a_1 a_2}{k (a_2 - a_1)} \]
Substituting \( k = \frac{1}{4 \pi \epsilon_0} \):
\[ Q_1 = 4 \pi \epsilon_0 (\phi_1 - \phi_2) \frac{a_1 a_2}{a_2 - a_1} \]

### Final Answer
Thus, the charge on the inner shell is:
\[ Q_1 = 4 \pi \epsilon_0 (\phi_1 - \phi_2) \frac{a_1 a_2}{a_2 - a_1} \]

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