A conducting sphere of radius R=20 cm is given a charge `Q=16muC`. What it `vecE` at centre

Options

A. `3.6xx10^(6)N//C`
B. `1.8xx10^(6)N//C`
C. Zero
D. `0.9xx10^(6)N//C`

Answer

To find the electric field \( \vec{E} \) at the center of a conducting sphere with a given charge, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Properties of a Conducting Sphere:**
- A conducting sphere allows charges to move freely on its surface. When a charge is placed on a conducting sphere, it distributes uniformly over the surface.

2. **Identify the Given Values:**
- Radius of the sphere, \( R = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Charge on the sphere, \( Q = 16 \, \mu\text{C} = 16 \times 10^{-6} \, \text{C} \)

3. **Electric Field Inside a Conductor:**
- Inside a conductor in electrostatic equilibrium, the electric field \( \vec{E} \) is always zero. This is because any electric field would cause the charges to move, contradicting the condition of electrostatic equilibrium.

4. **Applying Gauss's Law:**
- According to Gauss's Law, the electric field \( \vec{E} \) inside a closed surface is given by:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]
- For a point inside the conducting sphere (including the center), the charge enclosed \( Q_{\text{enc}} = 0 \) because all the charge resides on the surface.

5. **Conclusion:**
- Since \( Q_{\text{enc}} = 0 \), we have:
\[ \oint \vec{E} \cdot d\vec{A} = 0 \]
- This implies that the electric field \( \vec{E} \) at the center (and anywhere inside the conductor) is:
\[ \vec{E} = 0 \]

### Final Answer:
The electric field \( \vec{E} \) at the center of the conducting sphere is \( 0 \, \text{N/C} \).