A is a spherical conductor placed concentrically inside a hollow spherical conductor B. A is given +q charge and B is earthed. The n the electric intensity is not zero

Options

A. Inside A
B. Outside B
C. On the surface of B
D. Between A and B

Explanation

To solve the problem, we need to analyze the situation involving two spherical conductors, A and B, where A is a charged conductor and B is earthed. Here’s a step-by-step solution:

### Step 1: Understand the Configuration
- We have two spherical conductors: A (with charge +q) and B (hollow and earthed).
- Since A is inside B, they are concentric.

**Hint:** Visualize the arrangement of the two conductors and their charges.

### Step 2: Analyze the Charge Distribution
- When A is given a charge of +q, it induces a charge of -q on the inner surface of B (due to electrostatic induction).
- The outer surface of B will then have a charge of +q because B is earthed (which means it can exchange charge with the ground).

**Hint:** Remember that the total charge on a conductor must equal the charge it has after accounting for induced charges.

### Step 3: Apply Gauss's Law
- Inside conductor A, the electric field (E) is zero because the electric field inside a conductor in electrostatic equilibrium is always zero.
- For a Gaussian surface inside A, the enclosed charge (Q_in) is zero, leading to E = 0.

**Hint:** Use Gauss's law: \( \Phi = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{in}}{\epsilon_0} \).

### Step 4: Evaluate the Electric Field in Region Between A and B
- In the region between A and B, we can take a Gaussian surface that lies between the two conductors.
- The enclosed charge in this region is +q (the charge on A), which results in an electric field.

**Hint:** The electric field in this region can be calculated using \( E = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{r^2} \) where r is the distance from the center.

### Step 5: Check the Electric Field Outside B
- Outside conductor B, the total charge enclosed is zero (since the charges on A and the induced charges on B cancel out).
- Therefore, the electric field outside B is also zero.

**Hint:** Again apply Gauss's law to confirm that the net charge enclosed is zero.

### Conclusion
- The electric field is zero inside conductor A and outside conductor B.
- However, the electric field is not zero in the region between A and B, where the electric field can be calculated using the charge on A.

**Final Answer:** The electric intensity is not zero in the region between the two conductors (between A and B).