The electric field in a region is radially outward with magnitude `E=Ar`. Find the charge contained in a sphere of radius a centred at the origin. Take `A=100 V m^(-2)` and `a=20.0 cm`.

Explanation

To find the charge contained in a sphere of radius \( a \) centered at the origin, given that the electric field \( E \) in the region is radially outward with a magnitude of \( E = Ar \), we can follow these steps:

### Step 1: Identify the Electric Field
The electric field is given as:
\[ E = Ar \]
where \( A = 100 \, \text{V/m}^2 \) and \( r \) is the distance from the origin.

### Step 2: Determine the Electric Field at Radius \( a \)
For a sphere of radius \( a = 20.0 \, \text{cm} = 0.2 \, \text{m} \), we can substitute \( r = a \) into the electric field equation:
\[ E(a) = A \cdot a = 100 \cdot 0.2 = 20 \, \text{V/m} \]

### Step 3: Calculate the Surface Area of the Sphere
The surface area \( A_s \) of a sphere is given by the formula:
\[ A_s = 4\pi a^2 \]
Substituting \( a = 0.2 \, \text{m} \):
\[ A_s = 4\pi (0.2)^2 = 4\pi (0.04) = 0.50265 \, \text{m}^2 \]

### Step 4: Calculate the Electric Flux through the Sphere
The electric flux \( \Phi_E \) through the surface of the sphere is given by:
\[ \Phi_E = E \cdot A_s \]
Substituting the values we calculated:
\[ \Phi_E = 20 \cdot 0.50265 = 10.0529 \, \text{V m} \]

### Step 5: Use Gauss's Law to Find the Charge
According to Gauss's law, the total electric flux through a closed surface is equal to the charge enclosed \( Q \) divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi_E = \frac{Q}{\epsilon_0} \]
Rearranging for \( Q \):
\[ Q = \Phi_E \cdot \epsilon_0 \]
Where \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step 6: Substitute the Values to Find \( Q \)
Substituting the values:
\[ Q = 10.0529 \cdot (8.85 \times 10^{-12}) = 8.89 \times 10^{-11} \, \text{C} \]

### Final Result
The charge contained in the sphere of radius \( a \) is approximately:
\[ Q \approx 8.89 \times 10^{-11} \, \text{C} \]