A small conducting sphere of radius a mounted on an insulated handle and a positive charge q is inserted through a hole in the wall of a hollow conducting sphere of inner radius b and outer radius c. The hollow sphere is supported on an insulating stand and is initially uncharged. The small sphere is placed at the center of the hollow sphere. Neglect any effect of the hole. Which of the following statements will be true for this system?

Options

A. No work will be done in carrying a small charge from the inner conductor to the outer conductor.
B. The electric field at a point in the region between the spheres at a distance r from the center is equal to `q//4piepsilon_0r^2`.
C. The electric field at a point outside the hollow sphere at a distance r from the centre is `q//4piepsilon_0r^2`.
D. The potential of the inner sphere with respect to the outer sphere is given by `V_(ab) = q/(4piepsilon_0)(1/a-1/b)`.

Explanation

To solve the problem, we need to analyze the situation step by step, considering the properties of electric fields and potentials in conductors.

### Step 1: Understand the System
We have a small conducting sphere of radius \( a \) with a positive charge \( q \) placed at the center of a hollow conducting sphere with inner radius \( b \) and outer radius \( c \). The hollow sphere is initially uncharged and is insulated.

### Step 2: Charge Induction
When the small charged sphere is placed at the center of the hollow conducting sphere, it induces charges on the inner surface of the hollow sphere. The inner surface will acquire a negative charge of \( -q \) due to the presence of the positive charge \( q \) at the center. Consequently, the outer surface of the hollow sphere will acquire a positive charge of \( +q \) to keep the entire hollow sphere neutral.

### Step 3: Electric Field Inside the Hollow Sphere
To find the electric field at a point inside the hollow sphere (between the inner surface and the small sphere), we can apply Gauss's Law. The electric field \( E \) at a distance \( r \) from the center (where \( r < b \)) is given by:
\[ E \cdot 4\pi r^2 = \frac{q}{\epsilon_0} \]
Thus, the electric field \( E \) inside the hollow sphere is:
\[ E = \frac{q}{4\pi \epsilon_0 r^2} \]
This confirms that the electric field behaves as if all the charge \( q \) is concentrated at the center.

### Step 4: Electric Field Outside the Hollow Sphere
For points outside the hollow sphere (at a distance \( r \) from the center where \( r > c \)), we again apply Gauss's Law. The total charge enclosed by a Gaussian surface outside the hollow sphere is \( q \) (the positive charge on the outer surface). Thus, the electric field \( E \) at a distance \( r \) from the center is:
\[ E \cdot 4\pi r^2 = \frac{q}{\epsilon_0} \]
Therefore, the electric field outside the hollow sphere is:
\[ E = \frac{q}{4\pi \epsilon_0 r^2} \]

### Step 5: Potential Difference Between Inner and Outer Sphere
To find the potential difference between the inner sphere and the outer sphere, we calculate the potentials at both surfaces. The potential \( V_1 \) at the inner surface (radius \( b \)) is influenced by the charge \( q \) and the induced charge \( -q \):
\[ V_1 = k \left( \frac{q}{a} - \frac{q}{b} \right) \]
The potential \( V_2 \) at the outer surface (radius \( c \)) is:
\[ V_2 = k \left( \frac{q}{c} \right) \]
The potential difference \( V_1 - V_2 \) gives us:
\[ V_1 - V_2 = kq \left( \frac{1}{a} - \frac{1}{b} - \frac{1}{c} \right) \]

### Step 6: Work Done in Moving a Charge
The work done \( W \) in moving a small charge \( q' \) from the inner to the outer conductor is given by:
\[ W = q' \Delta V \]
Since \( \Delta V \) is not zero, work will be done in moving the charge.

### Conclusion
Based on the analysis:
- **Statement B**: True, the electric field inside the hollow sphere is \( \frac{q}{4\pi \epsilon_0 r^2} \).
- **Statement C**: True, the electric field outside the hollow sphere is also \( \frac{q}{4\pi \epsilon_0 r^2} \).
- **Statement D**: True, the potential difference is given by \( kq \left( \frac{1}{a} - \frac{1}{b} - \frac{1}{c} \right) \).
- **Statement A**: False, work will be done in moving a charge from the inner to the outer conductor.