A conducting spehre has a radius 30 cm .If the dielectric strength of surrounding air is `3'10^(6)` V/m, the maximum amount of charge the sphere can holdin micro coulombs is

Options

A. 0.03
B. 0.3
C. 3
D. 30

Explanation

To find the maximum amount of charge that a conducting sphere can hold, we can use the formula for the electric field around a charged sphere. The electric field \( E \) due to a point charge (or a charged sphere) at a distance \( r \) from the center is given by:

\[ E = \frac{k \cdot Q}{r^2} \]

where:
- \( E \) is the electric field strength,
- \( k \) is Coulomb's constant (\( k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( Q \) is the charge,
- \( r \) is the radius of the sphere.

### Step 1: Identify the given values
- Radius of the sphere \( r = 30 \, \text{cm} = 0.3 \, \text{m} \)
- Dielectric strength of air \( E = 3 \times 10^6 \, \text{V/m} \)

### Step 2: Rearrange the formula to solve for charge \( Q \)
From the formula for electric field, we can rearrange it to solve for \( Q \):

\[ Q = E \cdot r^2 / k \]

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation:

\[ Q = \frac{(3 \times 10^6) \cdot (0.3)^2}{9 \times 10^9} \]

### Step 4: Calculate \( r^2 \)
First, calculate \( r^2 \):

\[ (0.3)^2 = 0.09 \, \text{m}^2 \]

### Step 5: Substitute \( r^2 \) back into the equation
Now substituting \( r^2 \) back into the equation for \( Q \):

\[ Q = \frac{(3 \times 10^6) \cdot (0.09)}{9 \times 10^9} \]

### Step 6: Calculate the numerator
Calculating the numerator:

\[ 3 \times 10^6 \cdot 0.09 = 2.7 \times 10^6 \]

### Step 7: Divide by \( k \)
Now divide by \( k \):

\[ Q = \frac{2.7 \times 10^6}{9 \times 10^9} = 0.3 \times 10^{-3} \, \text{C} \]

### Step 8: Convert to microcoulombs
To convert to microcoulombs:

\[ Q = 0.3 \times 10^{-3} \, \text{C} = 30 \, \mu\text{C} \]

### Final Answer
The maximum amount of charge the sphere can hold is:

\[ \boxed{30 \, \mu\text{C}} \]