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

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 of calculating the electric field in terms of `q` and the distance `r` from the common center of two shells, we will analyze the system step by step.

### Step 1: Understand the Configuration
We have two concentric spherical shells:
- The inner shell has a charge of `+2q`.
- The outer shell has a charge of `-4q`.

### Step 2: Identify Regions of Interest
We need to analyze the electric field in different regions:
1. Inside the inner shell (r < radius of inner shell)
2. Between the inner and outer shells (radius of inner shell < r < radius of outer shell)
3. Outside the outer shell (r > radius of outer shell)

### Step 3: Electric Field Inside the Inner Shell (r < radius of inner shell)
According to Gauss's Law, the electric field inside a conductor is zero. Since the inner shell is a conductor and we are inside it, the electric field `E` is:
\[ E = 0 \]

### Step 4: Electric Field Between the Shells (radius of inner shell < r < radius of outer shell)
In this region, we will consider a Gaussian surface that is a sphere of radius `r` (where the radius is greater than that of the inner shell but less than that of the outer shell). The total charge enclosed by this Gaussian surface is the charge of the inner shell:
\[ Q_{\text{enc}} = +2q \]

Using Gauss's Law:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} = E \cdot 4\pi r^2 \]
Substituting for \( Q_{\text{enc}} \):
\[ E \cdot 4\pi r^2 = \frac{2q}{\epsilon_0} \]
Thus, the electric field \( E \) in this region is:
\[ E = \frac{2q}{4\pi \epsilon_0 r^2} = \frac{q}{2\pi \epsilon_0 r^2} \]

### Step 5: Electric Field Outside the Outer Shell (r > radius of outer shell)
In this region, the total charge enclosed by a Gaussian surface of radius `r` is the sum of the charges of both shells:
\[ Q_{\text{enc}} = +2q - 4q = -2q \]

Using Gauss's Law again:
\[ E \cdot 4\pi r^2 = \frac{-2q}{\epsilon_0} \]
Thus, the electric field \( E \) in this region is:
\[ E = \frac{-2q}{4\pi \epsilon_0 r^2} = \frac{-q}{2\pi \epsilon_0 r^2} \]

### Summary of Electric Field in Different Regions
1. For \( r < \) (radius of inner shell): \( E = 0 \)
2. For (radius of inner shell) < \( r < \) (radius of outer shell): \( E = \frac{q}{2\pi \epsilon_0 r^2} \)
3. For \( r > \) (radius of outer shell): \( E = \frac{-q}{2\pi \epsilon_0 r^2} \)