Question

Find work done in bringing charge q = 3nC from infinity to point A as shown in the figure : 

• A. \(11 \times 10^{-7}\) J
• B. \(36 \times 10^{-7}\) J
• C. \(12 \times 10^{-7}\) J
• D. \(13 \times 10^{-7}\) J

Hint:

Potential is a scalar. Simply sum the individual potentials.
Keep units consistent: cm must be converted to meters for standard \(k = 9 \times 10^9\).

Answer 1

The Correct Option is B

To find the work done in bringing a charge \( q = 3 \, \text{nC} \) from infinity to point A in the given configuration, we need to calculate the electric potential at point A and then use it to find the work done.

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

\(V = \frac{kQ}{r}\)

where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \) is Coulomb's constant.

In the triangle configuration, there are two charges affecting point A: 1 nC at B and 3 nC at C.

Step 1: Calculate the potential at A due to charge at B

Charge at B, \( Q_B = 1 \, \text{nC} = 1 \times 10^{-9} \, \text{C} \)

Distance \( AB = 3 \, \text{cm} = 0.03 \, \text{m} \)

\(V_B = \frac{k \times Q_B}{AB} = \frac{9 \times 10^9 \times 1 \times 10^{-9}}{0.03} = 300 \, \text{V}\)

Step 2: Calculate the potential at A due to charge at C

Charge at C, \( Q_C = 3 \, \text{nC} = 3 \times 10^{-9} \, \text{C} \)

Distance \( AC = 3 \, \text{cm} = 0.03 \, \text{m} \)

\(V_C = \frac{k \times Q_C}{AC} = \frac{9 \times 10^9 \times 3 \times 10^{-9}}{0.03} = 900 \, \text{V}\)

Step 3: Total potential at A

\(V_{\text{total}} = V_B + V_C = 300 + 900 = 1200 \, \text{V}\)

Step 4: Calculate work done in bringing charge to A

The work done \( W \) is given by:

\(W = q \times V_{\text{total}} = 3 \times 10^{-9} \times 1200 \, \text{J}\)

\(W = 3.6 \times 10^{-6} \, \text{J} = 36 \times 10^{-7} \, \text{J}\)

Therefore, the work done is \( 36 \times 10^{-7} \, \text{J} \).