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:

Options

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

Answer

To solve the problem of finding the electric field at a distance \( r \) from the center of three concentric spheres with surface charge densities \( \sigma \), \( -\sigma \), and \( \sigma \) on spheres of radii \( a \), \( b \), and \( c \) respectively (where \( a > b > c \)), we will use Gauss's Law.

### Step-by-Step Solution:

1. **Identify the Regions**:
We have three concentric spheres:
- Sphere 1 (radius \( c \)): surface charge density \( \sigma \)
- Sphere 2 (radius \( b \)): surface charge density \( -\sigma \)
- Sphere 3 (radius \( a \)): surface charge density \( \sigma \)

We are interested in the region where \( b > r > c \).

2. **Apply Gauss's Law**:
According to Gauss's Law, the electric field \( E \) through a closed surface is proportional to the charge enclosed by that surface:
\[ \Phi_E = \oint E \cdot dA = \frac{Q_{\text{enc}}}{\epsilon_0} \]
Here, \( dA \) is the differential area element on the Gaussian surface, and \( Q_{\text{enc}} \) is the total charge enclosed by the Gaussian surface.

3. **Choose a Gaussian Surface**:
We will take a spherical Gaussian surface of radius \( r \) (where \( b > r > c \)). The area of this surface is:
\[ A = 4\pi r^2 \]

4. **Calculate the Enclosed Charge \( Q_{\text{enc}} \)**:
The charge enclosed by the Gaussian surface includes the charge from the inner sphere (radius \( c \)) and excludes the charge from the outer sphere (radius \( b \)):
- Charge from sphere with radius \( c \):
\[ Q_c = \sigma \cdot 4\pi c^2 \]
- Charge from sphere with radius \( b \):
\[ Q_b = -\sigma \cdot 4\pi b^2 \]
Thus, the total enclosed charge is:
\[ Q_{\text{enc}} = Q_c + Q_b = 4\pi c^2 \sigma - 4\pi b^2 \sigma = 4\pi \sigma (c^2 - b^2) \]

5. **Substitute into Gauss's Law**:
Now substituting \( Q_{\text{enc}} \) into Gauss's Law:
\[ E \cdot 4\pi r^2 = \frac{4\pi \sigma (c^2 - b^2)}{\epsilon_0} \]

6. **Solve for Electric Field \( E \)**:
Dividing both sides by \( 4\pi r^2 \):
\[ E = \frac{\sigma (c^2 - b^2)}{\epsilon_0 r^2} \]

### Final Answer:
The magnitude of the electric field at a distance \( r \) from the center, where \( b > r > c \), is given by:
\[ E = \frac{\sigma (c^2 - b^2)}{\epsilon_0 r^2} \]