The electric field in a region is radially outward with magnitude `E=Ar_(0)` . The charge contained in a sphere of radius `r_(0)` centred at the origin is

Options

A. `(Ar_(0)^(3))/(4piepsilon_(0))`
B. `(4piepsilon_(0)A)/(r_(0))`
C. `4piepsilon_(0)Ar_(0)^(3)`
D. `(A)/(4pir^(3)epsilon_(0))`

Explanation

To find the charge contained in a sphere of radius \( r_0 \) centered at the origin, given that the electric field in the region is radially outward with a magnitude \( E = A r_0 \), we can use Gauss's law. Here’s a step-by-step solution:

### Step 1: Understand the Electric Field
The electric field \( E \) is given as \( E = A r_0 \), where \( A \) is a constant and \( r_0 \) is the radius of the sphere. The electric field is directed radially outward from the center of the sphere.

**Hint:** Remember that the electric field points away from positive charges and towards negative charges.

### Step 2: Apply Gauss's Law
According to Gauss's law, the electric flux \( \Phi_E \) through a closed surface is equal to the charge \( Q \) enclosed by that surface divided by the permittivity of free space \( \epsilon_0 \):

\[ \Phi_E = \frac{Q}{\epsilon_0} \]

### Step 3: Calculate the Electric Flux
The electric flux \( \Phi_E \) can also be expressed as the product of the electric field \( E \) and the area \( A \) through which it passes. For a sphere, the area \( A \) is given by:

\[ A = 4\pi r_0^2 \]

Thus, the electric flux is:

\[ \Phi_E = E \cdot A = E \cdot (4\pi r_0^2) \]

### Step 4: Substitute the Electric Field
Substituting the expression for \( E \):

\[ \Phi_E = (A r_0) \cdot (4\pi r_0^2) = 4\pi A r_0^3 \]

### Step 5: Relate Electric Flux to Charge
Now, equate the two expressions for electric flux:

\[ 4\pi A r_0^3 = \frac{Q}{\epsilon_0} \]

### Step 6: Solve for Charge \( Q \)
Rearranging the equation to solve for \( Q \):

\[ Q = 4\pi \epsilon_0 A r_0^3 \]

### Final Answer
The charge contained in the sphere of radius \( r_0 \) centered at the origin is:

\[ Q = 4\pi \epsilon_0 A r_0^3 \]

---