There are three concentric thin spheres of radius `a,b,c (agtbgtc)`. The total surface charge densities on their surfaces are `sigma,-sigma,sigma` respectively. The magnitude of the electric field at `r` (distance from centre) such that `agtrgtb` is:

A. 0
B. `(sigma)/(epsilon_(0) r^(2))(b^(2)-c^(2))`
C. `(sigma)/(epsilon_(0)r)(a^(2)+b^(2))`
D. none of these

Options

A. 0
B. `(sigma)/(epsilon_(0) r^(2))(b^(2)-c^(2))r^(3)`
C. `(sigma)/(epsilon_(0)r)(a^(2)+b^(2))`
D. none of these

Explanation

To solve the problem, we will use Gauss's law to find the electric field at a distance \( r \) from the center of the concentric spheres.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have three concentric spheres with radii \( a \), \( b \), and \( c \) such that \( a > b > c \).
- The surface charge densities are \( \sigma \) for the sphere of radius \( a \), \( -\sigma \) for the sphere of radius \( b \), and \( \sigma \) for the sphere of radius \( c \).

2. **Identifying the Region**:
- We need to find the electric field at a distance \( r \) from the center where \( b < r < a \).

3. **Applying Gauss's Law**:
- According to Gauss's law, the electric flux through a closed surface is equal to the charge enclosed divided by \( \epsilon_0 \):
\[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
- For a spherical Gaussian surface of radius \( r \), the area \( A \) is \( 4\pi r^2 \).

4. **Calculating the Enclosed Charge**:
- The charge enclosed by the Gaussian surface at radius \( r \) includes contributions from the spheres with radii \( c \) and \( b \):
- Charge from sphere of radius \( c \):
\[ Q_c = \sigma \cdot 4\pi c^2 \]
- Charge from sphere of radius \( b \):
\[ Q_b = -\sigma \cdot 4\pi b^2 \]
- Therefore, the total charge enclosed \( Q_{\text{enclosed}} \) is:
\[ Q_{\text{enclosed}} = Q_c + Q_b = \sigma \cdot 4\pi c^2 - \sigma \cdot 4\pi b^2 = 4\pi \sigma (c^2 - b^2) \]

5. **Finding the Electric Field**:
- Substitute \( Q_{\text{enclosed}} \) into Gauss's law:
\[ \oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi r^2 = \frac{4\pi \sigma (c^2 - b^2)}{\epsilon_0} \]
- Simplifying gives:
\[ E \cdot 4\pi r^2 = \frac{4\pi \sigma (c^2 - b^2)}{\epsilon_0} \]
- Therefore, the electric field \( E \) is:
\[ E = \frac{\sigma (c^2 - b^2)}{\epsilon_0 r^2} \]

6. **Final Result**:
- The magnitude of the electric field at distance \( r \) from the center is:
\[ E = \frac{\sigma (b^2 - c^2)}{\epsilon_0 r^2} \]
- Since \( b > c \), we can write it as:
\[ E = \frac{\sigma}{\epsilon_0 r^2} (b^2 - c^2) \]

### Conclusion:
The correct answer is option **B**:
\[ E = \frac{\sigma}{\epsilon_0 r^2} (b^2 - c^2) \]