The electric field in a region is radially outward and at a point is given by `E=250" r V"//"m"` (where r is the distance of the point from origin). Calculate the charge contained in a sphere of radius 20 cm centred at the origin

Options

A. `2.22xx10^(-6)C`
B. `2.22xx10^(-6)C`
C. `2.22xx10^(-10)C`
D. Zero

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Electric Field
The electric field \( E \) is given as:
\[ E = 250r \, \text{V/m} \]
where \( r \) is the distance from the origin. We need to calculate the charge contained in a sphere of radius \( r = 20 \, \text{cm} = 0.2 \, \text{m} \).

### Step 2: Calculate the Electric Field at the Surface of the Sphere
Substituting \( r = 0.2 \, \text{m} \) into the electric field equation:
\[ E = 250 \times 0.2 = 50 \, \text{V/m} \]

### Step 3: Use Gauss's Law
According to Gauss's Law, the electric flux \( \Phi_E \) through a closed surface is equal to the charge \( Q_{\text{in}} \) enclosed by that surface divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi_E = \frac{Q_{\text{in}}}{\epsilon_0} \]
The electric flux can also be expressed as:
\[ \Phi_E = E \cdot A \]
where \( A \) is the surface area of the sphere. The surface area \( A \) of a sphere is given by:
\[ A = 4\pi r^2 \]
Substituting \( r = 0.2 \, \text{m} \):
\[ A = 4\pi (0.2)^2 = 4\pi (0.04) = 0.16\pi \, \text{m}^2 \]

### Step 4: Calculate the Electric Flux
Now substituting the values of \( E \) and \( A \) into the flux equation:
\[ \Phi_E = E \cdot A = 50 \, \text{V/m} \cdot 0.16\pi \, \text{m}^2 = 8\pi \, \text{V m} \]

### Step 5: Relate Electric Flux to Charge
Now, using Gauss's Law:
\[ \frac{Q_{\text{in}}}{\epsilon_0} = 8\pi \]
Thus, we can find \( Q_{\text{in}} \):
\[ Q_{\text{in}} = 8\pi \epsilon_0 \]
Using the value of \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \):
\[ Q_{\text{in}} = 8\pi (8.85 \times 10^{-12}) \approx 2.22 \times 10^{-10} \, \text{C} \]

### Final Answer
The charge contained in the sphere of radius 20 cm is approximately:
\[ Q_{\text{in}} \approx 2.22 \times 10^{-10} \, \text{C} \]