A `5C` charge experience a force of `2000 N` when moved between two points separated by a distance of `2 cm` in a uniform electric field. The potential difference between the two points is

Options

A. 8 V
B. 200 V
C. 800 V
D. 20, 000 V

Answer

To find the potential difference between two points in a uniform electric field, we can use the relationship between electric field (E), force (F), charge (q), and potential difference (V). Here’s the step-by-step solution:

### Step 1: Identify the given values
- Charge (q) = 5 C
- Force (F) = 2000 N
- Distance (d) = 2 cm = 0.02 m (conversion from cm to m)

### Step 2: Calculate the electric field (E)
The electric field can be calculated using the formula:
\[ E = \frac{F}{q} \]
Substituting the values:
\[ E = \frac{2000 \, \text{N}}{5 \, \text{C}} = 400 \, \text{N/C} \]

### Step 3: Calculate the potential difference (V)
The potential difference between two points in an electric field is given by:
\[ V = E \cdot d \]
Substituting the values we have:
\[ V = 400 \, \text{N/C} \times 0.02 \, \text{m} \]

### Step 4: Perform the multiplication
\[ V = 400 \times 0.02 = 8 \, \text{V} \]

### Conclusion
The potential difference between the two points is **8 V**.