Two points A and B are 2 cm apart and a uniform electric field E acts along the straight line `AB` directed A to B with `E=200N//C`. A particle of charge `+10^-6C` is taken from A to B along AB, Calculate
a. the force on the charge
b. the potential difference `V_A -V_B` and
c.the work done on the charge by E

Explanation

To solve the problem step by step, we will calculate the force on the charge, the potential difference \( V_A - V_B \), and the work done on the charge by the electric field \( E \).

### Given Data:
- Distance between points A and B, \( d = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Electric field, \( E = 200 \, \text{N/C} \)
- Charge, \( q = +10^{-6} \, \text{C} \)

### Part (a): Calculate the Force on the Charge
The force \( F \) on a charge in an electric field is given by the formula:
\[ F = q \cdot E \]
Substituting the values:
\[ F = (10^{-6} \, \text{C}) \cdot (200 \, \text{N/C}) = 2 \times 10^{-4} \, \text{N} \]

### Part (b): Calculate the Potential Difference \( V_A - V_B \)
The potential difference \( V_A - V_B \) in a uniform electric field is given by:
\[ V_A - V_B = -E \cdot d \]
Substituting the values:
\[ V_A - V_B = - (200 \, \text{N/C}) \cdot (0.02 \, \text{m}) = -4 \, \text{V} \]
Thus, the magnitude of the potential difference is \( 4 \, \text{V} \).

### Part (c): Calculate the Work Done on the Charge by \( E \)
The work done \( W \) on the charge by the electric field is given by:
\[ W = q \cdot (V_A - V_B) \]
Substituting the values:
\[ W = (10^{-6} \, \text{C}) \cdot (-4 \, \text{V}) = -4 \times 10^{-6} \, \text{J} \]

### Summary of Results:
- **Force on the charge**: \( 2 \times 10^{-4} \, \text{N} \)
- **Potential difference \( V_A - V_B \)**: \( -4 \, \text{V} \) (or \( 4 \, \text{V} \) in magnitude)
- **Work done on the charge by \( E \)**: \( -4 \times 10^{-6} \, \text{J} \)