When a charge of `3` coulombs is placed in a uniform electric field, it experiences a force of `3000` netwon. Within this field, potential difference between two points separated by a distance of `1 cm` is

Options

A. 10 V
B. 100 V
C. 30 V
D. 300 V

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Electric Field (E)
We know that the electric field (E) can be calculated using the formula:
\[ E = \frac{F}{q} \]
where:
- \( F \) is the force experienced by the charge,
- \( q \) is the charge.

Given:
- \( F = 3000 \, \text{N} \)
- \( q = 3 \, \text{C} \)

Substituting the values:
\[ E = \frac{3000 \, \text{N}}{3 \, \text{C}} = 1000 \, \text{N/C} \]

### Step 2: Calculate the Potential Difference (V)
The potential difference (V) between two points in an electric field can be calculated using the formula:
\[ V = E \times d \]
where:
- \( d \) is the distance between the two points.

Given:
- \( d = 1 \, \text{cm} = 0.01 \, \text{m} \) (convert centimeters to meters)

Substituting the values:
\[ V = 1000 \, \text{N/C} \times 0.01 \, \text{m} = 10 \, \text{V} \]

### Final Answer
The potential difference between the two points is \( 10 \, \text{V} \).

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