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
`a lt r lt b`

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 need to calculate the electric field in the region between the inner shell and the outer shell, specifically for the distance \( r \) where \( a < r < b \).

### Step-by-Step Solution:

1. **Identify the Charges**:
- The inner shell has a charge of \( +2q \).
- The outer shell has a charge of \( -4q \).

2. **Understand the Region**:
- We are interested in the region where \( a < r < b \). This means we are looking at points that are outside the inner shell but inside the outer shell.

3. **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, \( \Phi_E \) is the electric flux, \( E \) is the electric field, \( dA \) is the differential area, and \( Q_{\text{enc}} \) is the charge enclosed by the Gaussian surface.

4. **Choose a Gaussian Surface**:
- We will choose a spherical Gaussian surface of radius \( r \) such that \( a < r < b \).

5. **Calculate the Enclosed Charge**:
- In this region, the only charge enclosed by our Gaussian surface is the charge of the inner shell, which is \( +2q \). The charge on the outer shell does not contribute to the enclosed charge because it is outside our Gaussian surface.

6. **Substitute into Gauss's Law**:
- Since the only charge enclosed is \( +2q \), we can write:
\[ \oint E \cdot dA = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{2q}{\epsilon_0} \]
- The left side can be expressed as \( E \cdot 4\pi r^2 \) (the surface area of a sphere):
\[ E \cdot 4\pi r^2 = \frac{2q}{\epsilon_0} \]

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

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