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 of radius 10 cm, given that the electric field at a distance of 15 cm from the center of the sphere is \(3 \times 10^3\) N/C and directed inward, we can follow these steps:

### Step 1: Understand the Problem
We have a conducting sphere with a radius of 10 cm. We need to find the net charge on this sphere based on the electric field measured at a distance of 15 cm from the center of the sphere.

### Step 2: Use Gauss's Law
Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (\(\epsilon_0\)):
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
The electric flux (\(\Phi_E\)) can also be expressed as:
\[ \Phi_E = E \cdot A \]
where \(E\) is the electric field and \(A\) is the area of the Gaussian surface.

### Step 3: Define the Gaussian Surface
We will consider a spherical Gaussian surface of radius 15 cm (0.15 m) surrounding the conducting sphere. The area \(A\) of a sphere is given by:
\[ A = 4\pi r^2 \]
Substituting \(r = 0.15\) m:
\[ A = 4\pi (0.15)^2 = 4\pi (0.0225) \approx 0.2827 \, \text{m}^2 \]

### Step 4: Calculate the Charge Enclosed
From Gauss's Law, we can equate the expressions for electric flux:
\[ \frac{Q_{\text{enc}}}{\epsilon_0} = E \cdot A \]
Substituting the values:
- \(E = 3 \times 10^3 \, \text{N/C}\)
- \(\epsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\)

Now we can calculate \(Q_{\text{enc}}\):
\[ Q_{\text{enc}} = E \cdot A \cdot \epsilon_0 \]
Substituting the values:
\[ Q_{\text{enc}} = (3 \times 10^3) \cdot (0.2827) \cdot (8.85 \times 10^{-12}) \]
Calculating this gives:
\[ Q_{\text{enc}} \approx 3 \times 10^3 \cdot 0.2827 \cdot 8.85 \times 10^{-12} \approx 7.5 \times 10^{-9} \, \text{C} \]

### Step 5: Determine the Sign of the Charge
Since the electric field is directed inward, this indicates that the charge on the sphere must be negative. Therefore, the net charge on the conducting sphere is:
\[ Q = -7.5 \times 10^{-9} \, \text{C} \]

### Final Answer
The net charge on the conducting sphere is:
\[ \boxed{-7.5 \times 10^{-9} \, \text{C}} \]