The electric field in a region is radially outward with magnitude E = 2r. The charge contained in a sphere of radius a = 2m centerd at the origin is `4x piepsi_(0)`. Find the value of x.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Electric Field
The electric field \( E \) is given as \( E = 2r \), where \( r \) is the distance from the origin. This electric field is radially outward.

### Step 2: Calculate the Electric Flux
The electric flux \( \Phi_E \) through a surface is given by the formula:
\[ \Phi_E = E \cdot A \]
where \( A \) is the area of the surface. For a sphere of radius \( a \), the area \( A \) is:
\[ A = 4\pi a^2 \]
Substituting the expression for the electric field, we have:
\[ \Phi_E = E \cdot A = (2r) \cdot (4\pi r^2) \]
Since we are considering a sphere of radius \( a = 2 \, \text{m} \), we substitute \( r = 2 \):
\[ \Phi_E = (2 \cdot 2) \cdot (4\pi (2)^2) = 4 \cdot (4\pi \cdot 4) = 16\pi \]

### Step 3: Apply Gauss's Law
According to Gauss's law, the electric flux through a closed surface is equal to the charge \( Q \) enclosed divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi_E = \frac{Q}{\epsilon_0} \]
Setting the two expressions for electric flux equal to each other, we have:
\[ 16\pi = \frac{Q}{\epsilon_0} \]

### Step 4: Relate Charge to Given Expression
We are given that the charge contained in the sphere is \( Q = 4x\pi\epsilon_0 \). Substituting this into the equation gives:
\[ 16\pi = \frac{4x\pi}{\epsilon_0} \]

### Step 5: Solve for \( x \)
To find \( x \), we can rearrange the equation:
\[ Q = 16\pi\epsilon_0 \]
Equating the two expressions for \( Q \):
\[ 4x\pi = 16\pi\epsilon_0 \]
Dividing both sides by \( 4\pi \):
\[ x = 4 \]

### Final Answer
The value of \( x \) is \( 4 \).

---