What is the net charge on a conducting sphere of radius 10 cm ? Given that the electric field 15 cm from the center of the sphere is equal to `3xx10^(3)` N/C and is directed inward

Options

A. `-7.5xx10^(-5)` C
B. `-7.5xx10^(-9)` C
C. `7.5xx10^(-5)` C
D. `7.5xx10^(-9)` C

Answer

To find the net charge on a conducting sphere based on the given electric field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of the sphere, \( R = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Distance from the center where the electric field is measured, \( r = 15 \, \text{cm} = 0.15 \, \text{m} \)
- Electric field, \( E = 3 \times 10^3 \, \text{N/C} \) (directed inward)

2. **Understand the Concept of Electric Field:**
The electric field \( E \) outside a conducting sphere is given by the formula:
\[ E = \frac{k \cdot Q}{r^2} \]
where \( k = \frac{1}{4\pi \epsilon_0} \) (Coulomb's constant), \( Q \) is the net charge on the sphere, and \( r \) is the distance from the center of the sphere.

3. **Use the Value of \( k \):**
The value of \( k \) is approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

4. **Set Up the Equation:**
Since the electric field is directed inward, it indicates that the charge on the sphere is negative. We can set up the equation as follows:
\[ 3 \times 10^3 = \frac{9 \times 10^9 \cdot Q}{(0.15)^2} \]

5. **Calculate \( (0.15)^2 \):**
\[ (0.15)^2 = 0.0225 \]

6. **Substitute and Rearrange the Equation:**
\[ 3 \times 10^3 = \frac{9 \times 10^9 \cdot Q}{0.0225} \]
Rearranging gives:
\[ Q = \frac{3 \times 10^3 \cdot 0.0225}{9 \times 10^9} \]

7. **Calculate \( Q \):**
\[ Q = \frac{67.5 \times 10^{-6}}{9 \times 10^9} = 7.5 \times 10^{-9} \, \text{C} \]

8. **Determine the Sign of the Charge:**
Since the electric field is directed inward, the charge is negative:
\[ Q = -7.5 \times 10^{-9} \, \text{C} \text{ or } -7.5 \, \text{nC} \]

### Final Answer:
The net charge on the conducting sphere is \( -7.5 \, \text{nC} \).

---