A charge of `5 C` experiences a force of `5000N` when it is kept in a uniform electric field. What is the potential difference between two points separated by a distance of `1 cm` ?

Options

A. `10 V`
B. `250 V`
C. `1000 V`
D. `2500 V`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between force, charge, and electric field
The force \( F \) experienced by a charge \( q \) in an electric field \( E \) is given by the formula:
\[ F = q \cdot E \]
Where:
- \( F \) is the force in newtons (N)
- \( q \) is the charge in coulombs (C)
- \( E \) is the electric field in newtons per coulomb (N/C)

### Step 2: Calculate the electric field
We can rearrange the formula to solve for the electric field \( E \):
\[ E = \frac{F}{q} \]
Substituting the given values:
- \( F = 5000 \, \text{N} \)
- \( q = 5 \, \text{C} \)

Calculating \( E \):
\[ E = \frac{5000 \, \text{N}}{5 \, \text{C}} = 1000 \, \text{N/C} \]

### Step 3: Understand the relationship between electric field and potential difference
The potential difference \( V \) between two points in a uniform electric field is given by:
\[ V = E \cdot d \]
Where:
- \( V \) is the potential difference in volts (V)
- \( E \) is the electric field in newtons per coulomb (N/C)
- \( d \) is the distance in meters (m)

### Step 4: Convert distance from centimeters to meters
The distance given is \( 1 \, \text{cm} \). We need to convert this to meters:
\[ d = 1 \, \text{cm} = 0.01 \, \text{m} \]

### Step 5: Calculate the potential difference
Now we can substitute the values into the formula for potential difference:
\[ V = E \cdot d = 1000 \, \text{N/C} \cdot 0.01 \, \text{m} \]
Calculating \( V \):
\[ V = 10 \, \text{V} \]

### Final Answer
The potential difference between the two points separated by a distance of 1 cm is \( 10 \, \text{V} \).

---