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
The graph of the radial component of `E` as function of `r` will be

Options

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B.
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D.

Explanation

To solve the problem, we need to calculate the electric field \( E \) at different regions defined by the distances \( r \) from the common center of the two shells. We have two spherical shells: an inner shell with charge \( +2q \) and an outer shell with charge \( -4q \).

### Step-by-Step Solution:

1. **Identify the Regions**:
We will analyze the electric field in three distinct regions:
- Region I: Inside the inner shell ( \( r < R_1 \) )
- Region II: Between the inner and outer shells ( \( R_1 < r < R_2 \) )
- Region III: Outside the outer shell ( \( r > R_2 \) )

2. **Region I: Inside the Inner Shell ( \( r < R_1 \) )**:
- By Gauss's law, the electric field inside a conductor or a shell with no enclosed charge is zero.
- Therefore, \( E = 0 \) for \( r < R_1 \).

3. **Region II: Between the Shells ( \( R_1 < r < R_2 \) )**:
- Here, we only consider the charge of the inner shell, since the outer shell's charge does not affect the electric field in this region.
- The enclosed charge \( Q_{\text{enc}} = +2q \).
- Applying Gauss's law:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]
The area \( A = 4\pi r^2 \), thus:
\[ E \cdot 4\pi r^2 = \frac{2q}{\epsilon_0} \]
Hence,
\[ E = \frac{2q}{4\pi \epsilon_0 r^2} = \frac{q}{2\pi \epsilon_0 r^2} \]

4. **Region III: Outside the Outer Shell ( \( r > R_2 \) )**:
- In this region, we consider the total enclosed charge, which is the sum of the charges of both shells.
- The total enclosed charge \( Q_{\text{enc}} = +2q - 4q = -2q \).
- Again applying Gauss's law:
\[ E \cdot 4\pi r^2 = \frac{-2q}{\epsilon_0} \]
Therefore,
\[ E = \frac{-2q}{4\pi \epsilon_0 r^2} = \frac{-q}{2\pi \epsilon_0 r^2} \]

### Summary of Electric Field in Different Regions:
- For \( r < R_1 \): \( E = 0 \)
- For \( R_1 < r < R_2 \): \( E = \frac{q}{2\pi \epsilon_0 r^2} \)
- For \( r > R_2 \): \( E = \frac{-q}{2\pi \epsilon_0 r^2} \)