A total charge of `5 muC` is distributed uniformly on the surface of the thin walled semispherical cup. If the electric field strength at the centre of the hemisphere is `9xx10^(8) NC^(-1)`, then the radius of the cup is
`(1/(4pi epsi_(0))=9xx10^(9) N-m^(2) C^(-2))`

Options

A. 5 mm
B. 10 mm
C. 5 cm
D. 10 cm

Answer

To solve the problem step by step, we will use the relationship between electric field, charge, and radius for a hemispherical cup.

### Step 1: Understand the given data
We have:
- Total charge \( Q = 5 \, \mu C = 5 \times 10^{-6} \, C \)
- Electric field strength at the center \( E = 9 \times 10^{8} \, N/C \)
- The constant \( \frac{1}{4\pi \epsilon_0} = 9 \times 10^{9} \, N \cdot m^{2}/C^{2} \)

### Step 2: Use the formula for electric field at the center of a hemispherical cup
The electric field \( E \) at the center of a hemispherical cup can be expressed as:
\[ E = \frac{\sigma}{4 \epsilon_0} \]
where \( \sigma \) is the surface charge density.

### Step 3: Calculate the surface charge density \( \sigma \)
The surface charge density \( \sigma \) is given by:
\[ \sigma = \frac{Q}{A} \]
For a hemisphere, the area \( A \) is:
\[ A = 2\pi r^2 \]
Thus,
\[ \sigma = \frac{Q}{2\pi r^2} \]

### Step 4: Substitute \( \sigma \) in the electric field equation
Substituting \( \sigma \) into the electric field equation gives:
\[ E = \frac{Q}{2\pi r^2 \cdot 4 \epsilon_0} = \frac{Q}{8\pi \epsilon_0 r^2} \]

### Step 5: Rearrange the equation to find \( r^2 \)
Rearranging the equation for \( r^2 \):
\[ E = \frac{Q}{8\pi \epsilon_0 r^2} \implies r^2 = \frac{Q}{8\pi \epsilon_0 E} \]

### Step 6: Substitute the known values
Substituting \( Q = 5 \times 10^{-6} \, C \), \( \epsilon_0 = \frac{1}{4\pi \cdot 9 \times 10^{9}} \), and \( E = 9 \times 10^{8} \, N/C \):
\[ r^2 = \frac{5 \times 10^{-6}}{8\pi \cdot 9 \times 10^{9} \cdot 9 \times 10^{8}} \]

### Step 7: Calculate the value
First, calculate \( 8\pi \cdot 9 \times 10^{9} \):
\[ 8\pi \cdot 9 \approx 226.194 \times 10^{9} \]
Now substituting:
\[ r^2 = \frac{5 \times 10^{-6}}{226.194 \times 10^{9} \cdot 9 \times 10^{8}} = \frac{5 \times 10^{-6}}{2.025 \times 10^{19}} \approx 2.47 \times 10^{-25} \]

### Step 8: Find \( r \)
Taking the square root:
\[ r = \sqrt{2.47 \times 10^{-25}} \approx 0.5 \times 10^{-2} \, m = 5 \, mm \]

### Final Answer
The radius of the cup is \( 5 \, mm \).

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