The potential at a point due to charge of `5xx10^(-7)C` located 10 cm away is
In the above question work done in bringing a charge of `4 xx 10^(-9)C` from infinity to that point is

Options

A. `2.4 xx 10^(-4)J`
B. `1.8 xx 10^(-4)J`
C. `3.2 xx 10^(-5)J`
D. `4.1 xx 10^(-5)J`

Explanation

To solve the problem step by step, we will first calculate the electric potential at a point due to a charge and then use that potential to find the work done in bringing another charge from infinity to that point.

### Step 1: Calculate the Electric Potential (V) at a distance of 10 cm

The formula for the electric potential \( V \) due to a point charge \( Q \) at a distance \( r \) is given by:

\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{r} \]

Where:
- \( Q = 5 \times 10^{-7} \, \text{C} \) (the charge)
- \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)

Substituting the values into the formula:

\[ V = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{5 \times 10^{-7}}{0.1} \]

Calculating \( \frac{1}{4 \pi \epsilon_0} \):

\[ \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \]

Now substituting this into the equation for \( V \):

\[ V = 9 \times 10^9 \cdot \frac{5 \times 10^{-7}}{0.1} = 9 \times 10^9 \cdot 5 \times 10^{-6} = 45 \times 10^3 \, \text{V} = 4.5 \times 10^4 \, \text{V} \]

### Step 2: Calculate the Work Done (W) in bringing a charge from infinity to that point

The work done \( W \) in bringing a charge \( q \) from infinity to a point with potential \( V \) is given by:

\[ W = q \cdot V \]

Where:
- \( q = 4 \times 10^{-9} \, \text{C} \)
- \( V = 4.5 \times 10^4 \, \text{V} \)

Substituting the values:

\[ W = (4 \times 10^{-9}) \cdot (4.5 \times 10^4) \]

Calculating \( W \):

\[ W = 18 \times 10^{-5} \, \text{J} = 1.8 \times 10^{-4} \, \text{J} \]

### Final Answer
The work done in bringing a charge of \( 4 \times 10^{-9} \, \text{C} \) from infinity to that point is \( 1.8 \times 10^{-4} \, \text{J} \).

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