Two concentric coducting thin spherical shells A and B having radii `r_A` and `r_B(r_Bgtr_A)` are charged to `Q_A` and `-Q_B(|Q_B|gt|Q_A|)`. The electrical field along a line passing through the centre is

Options

A.
B.
C.
D. none of these

Explanation

To solve the problem of finding the electric field along a line passing through the center of two concentric conducting spherical shells A and B, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have two concentric conducting spherical shells: Shell A with radius \( r_A \) and charge \( Q_A \), and Shell B with radius \( r_B \) (where \( r_B > r_A \)) and charge \( -Q_B \) (where \( |Q_B| > |Q_A| \)).
- The electric field inside a conductor in electrostatic equilibrium is zero.

2. **Electric Field Inside Shell A (0 to \( r_A \))**:
- For any point inside Shell A (i.e., at a distance \( r < r_A \)), the electric field \( E = 0 \) because the electric field inside a conducting shell is always zero.

3. **Electric Field Between Shells (from \( r_A \) to \( r_B \))**:
- In the region between the two shells (i.e., \( r_A < r < r_B \)), we can use Gauss's Law. The electric field is due to the charge on Shell A.
- The effective charge enclosed by a Gaussian surface in this region is \( Q_A \).
- According to Gauss's Law:
\[ E \cdot 4\pi r^2 = \frac{Q_A}{\varepsilon_0} \]
- Therefore, the electric field \( E \) at a distance \( r \) from the center is given by:
\[ E = \frac{Q_A}{4\pi \varepsilon_0 r^2} \]
- This electric field is directed outward because \( Q_A \) is positive.

4. **Electric Field Outside Shell B (from \( r > r_B \))**:
- For points outside Shell B (i.e., \( r > r_B \)), we consider the net charge enclosed by a Gaussian surface that encompasses both shells.
- The total charge enclosed is \( Q_A - Q_B \).
- Again applying Gauss's Law:
\[ E \cdot 4\pi r^2 = \frac{Q_A - Q_B}{\varepsilon_0} \]
- Thus, the electric field \( E \) for \( r > r_B \) is:
\[ E = \frac{Q_A - Q_B}{4\pi \varepsilon_0 r^2} \]
- Since \( |Q_B| > |Q_A| \), \( Q_A - Q_B < 0 \), indicating that the electric field is directed inward.

5. **Summary of Electric Field Behavior**:
- \( E = 0 \) for \( 0 < r < r_A \)
- \( E = \frac{Q_A}{4\pi \varepsilon_0 r^2} \) for \( r_A < r < r_B \) (outward)
- \( E = \frac{Q_A - Q_B}{4\pi \varepsilon_0 r^2} \) for \( r > r_B \) (inward)