A sphere of mass 50 gm is suspended by a string in an electric field of intensity `5NC^(-1)` acting vertically upward. If the tension in the string is 520 millinewton, the charge on the sphere is `(g = 10ms^(-2))`

Options

A. `-4 xx 10^(-3)C`
B. `4 xx 10^(3)C`
C. `8 xx 10^(3) C`
D. `-8 xx 10^(-3) C`

Answer

To solve the problem, we need to analyze the forces acting on the sphere suspended in the electric field.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the sphere, \( m = 50 \, \text{g} = 0.05 \, \text{kg} \)
- Electric field intensity, \( E = 5 \, \text{N/C} \)
- Tension in the string, \( T = 520 \, \text{mN} = 520 \times 10^{-3} \, \text{N} = 0.52 \, \text{N} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Weight of the Sphere:**
\[ \text{Weight} (W) = m \cdot g = 0.05 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 0.5 \, \text{N} \]

3. **Set Up the Force Equation:**
- The forces acting on the sphere are the tension \( T \) acting upwards and the weight \( W \) acting downwards.
- The electric force \( F_e \) acting on the sphere due to the electric field is given by \( F_e = qE \), where \( q \) is the charge on the sphere.
- According to the equilibrium of forces:
\[ T + F_e = W \]
Substituting \( F_e = qE \):
\[ T + qE = W \]

4. **Substitute the Known Values:**
\[ 0.52 \, \text{N} + q(5 \, \text{N/C}) = 0.5 \, \text{N} \]

5. **Rearranging the Equation:**
\[ q(5) = 0.5 - 0.52 \]
\[ q(5) = -0.02 \]

6. **Solve for Charge \( q \):**
\[ q = \frac{-0.02}{5} = -0.004 \, \text{C} \]
\[ q = -4 \times 10^{-3} \, \text{C} \]

### Final Answer:
The charge on the sphere is \( q = -4 \, \text{mC} \) (milliCoulombs).

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