Consider the previous problem , let the outer shell have the charge ` - 4 q`. As in the above problem , the inner shell has the charge ` + 2q`. Calculate the electric field in terms of `q` and the distance `r` from the common center of the two shells for
`b lt r lt c`

Options

A. zero
B. `(1)/( 4 pi epsilon_(0)) ( 2q)/( r^(2))`
C. `-(1)/( 4 pi epsilon_(0)) ( 2q)/( r^(2))`
D. `(1)/( 4 pi epsilon_(0)) ( 6 q)/( r^(2))`

Explanation

To solve the problem, we will use Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε₀).

### Step-by-Step Solution:

1. **Identify the Region of Interest**:
We need to find the electric field in the region where \( b < r < c \). This means we are looking at a point located between the inner shell and the outer shell.

2. **Determine the Charges**:
- The inner shell has a charge of \( +2q \).
- The outer shell has a charge of \( -4q \).
- The inner surface of the outer shell will have an induced charge of \( -2q \) due to the presence of the \( +2q \) charge on the inner shell.

3. **Construct a Gaussian Surface**:
We will consider a spherical Gaussian surface of radius \( r \) such that \( b < r < c \). This surface will be concentric with the shells.

4. **Calculate the Charge Enclosed**:
The charge enclosed by our Gaussian surface is only the charge of the inner shell, which is \( +2q \). The induced charge on the inner surface of the outer shell does not contribute to the enclosed charge since it is outside our Gaussian surface.

5. **Apply Gauss's Law**:
According to Gauss's Law:
\[ \Phi_E = \frac{Q_{enc}}{\epsilon_0} \]
where \( \Phi_E \) is the electric flux and \( Q_{enc} \) is the charge enclosed.

The electric flux \( \Phi_E \) can also be expressed as:
\[ \Phi_E = E \cdot A \]
where \( A \) is the area of the Gaussian surface. For a sphere, \( A = 4\pi r^2 \).

6. **Set Up the Equation**:
Thus, we have:
\[ E \cdot 4\pi r^2 = \frac{2q}{\epsilon_0} \]

7. **Solve for Electric Field \( E \)**:
Rearranging the equation gives:
\[ E = \frac{2q}{4\pi \epsilon_0 r^2} \]
Simplifying this, we get:
\[ E = \frac{q}{2\pi \epsilon_0 r^2} \]

### Final Answer:
The electric field \( E \) in the region where \( b < r < c \) is:
\[ E = \frac{q}{2\pi \epsilon_0 r^2} \]