A positively charged ball hangs from a long silk thread. Electric filed at a certain point (at the same horizontal level of ball) due to this charge is E. Let us put a positive test charge `q_(0)` at this point and measure `F//q_(0)` on this charges. then E

Options

A. `gtF//q_(0)`
B. `ltF//q_(0)`
C. `=F//q_(0)`
D. none of these

Explanation

To solve the problem, we need to analyze the situation of a positively charged ball hanging from a silk thread and the effect of placing a positive test charge \( q_0 \) at a certain point in the electric field created by the charged ball.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a positively charged ball with charge \( Q \) hanging from a silk thread.
- The electric field \( E \) at a certain horizontal point \( P \) (at the same level as the ball) due to this charge is given by the formula:
\[ E = \frac{kQ}{r^2} \]
where \( k \) is Coulomb's constant and \( r \) is the distance from the charge to point \( P \).

**Hint**: Remember that the electric field due to a point charge is directed away from the charge if it is positive.

2. **Placing the Test Charge**:
- We place a positive test charge \( q_0 \) at point \( P \).
- The force \( F \) experienced by the test charge due to the electric field \( E \) is given by:
\[ F = q_0 E \]

**Hint**: The force on a charge in an electric field is the product of the charge and the electric field strength.

3. **Calculating the Electric Field with the Test Charge**:
- When we place the test charge \( q_0 \), it will experience a repulsive force due to the charged ball. This will cause the silk thread to move, changing the distance between the ball and the point \( P \) to a new distance \( r' \).
- The new electric field \( E' \) at point \( P \) (after placing \( q_0 \)) can be expressed as:
\[ E' = \frac{F}{q_0} = \frac{kQ}{r'^2} \]

**Hint**: The electric field is still defined as the force per unit charge, but now the distance has changed.

4. **Comparing Distances**:
- Since the test charge \( q_0 \) is repelled by the positively charged ball, the new distance \( r' \) will be greater than the original distance \( r \) (i.e., \( r' > r \)).
- Therefore, we can conclude that:
\[ E' = \frac{kQ}{r'^2} < \frac{kQ}{r^2} = E \]

**Hint**: The electric field decreases with increasing distance from the charge.

5. **Final Conclusion**:
- Since \( E' < E \), we can state that:
\[ E > \frac{F}{q_0} \]
- Thus, the electric field \( E \) at point \( P \) is greater than the force per unit charge \( \frac{F}{q_0} \).

**Hint**: This relationship shows how the presence of the test charge affects the electric field and the forces involved.

### Final Answer:
\[ E > \frac{F}{q_0} \]