A positive charge of `17.7 muC` is placed at the centre of a hollow sphere of radius `0.5m`. Calculate the flux density through the surface of the sphere.

Explanation

To solve the problem of calculating the electric flux density through the surface of a hollow sphere with a charge at its center, we can follow these steps:

### Step 1: Understand the given data
We have a positive charge \( Q = 17.7 \, \mu C = 17.7 \times 10^{-6} \, C \) placed at the center of a hollow sphere with a radius \( r = 0.5 \, m \).

### Step 2: Use Gauss's Law
According to Gauss's Law, the electric flux \( \Phi \) through a closed surface is given by:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( Q_{\text{enc}} \) is the charge enclosed by the surface, and \( \epsilon_0 \) is the permittivity of free space, which has a value of \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \).

### Step 3: Calculate the electric flux
Substituting the values into Gauss's Law:
\[ \Phi = \frac{17.7 \times 10^{-6}}{8.85 \times 10^{-12}} \]
Calculating this gives:
\[ \Phi \approx 1.997 \times 10^{6} \, N \cdot m^2/C \]

### Step 4: Calculate the surface area of the sphere
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4 \pi r^2 \]
Substituting the radius \( r = 0.5 \, m \):
\[ A = 4 \pi (0.5)^2 = 4 \pi (0.25) = \pi \, m^2 \]

### Step 5: Calculate the electric flux density
The electric flux density \( D \) is defined as:
\[ D = \frac{\Phi}{A} \]
Substituting the values of \( \Phi \) and \( A \):
\[ D = \frac{1.997 \times 10^{6}}{\pi} \]
Calculating this gives:
\[ D \approx 636,000 \, N/C \quad \text{(approximately } 6.36 \times 10^5 \, N/C\text{)} \]

### Final Answer:
The electric flux density through the surface of the sphere is approximately \( 6.36 \times 10^5 \, N/C \).

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