Three particles, each having a charge of 10 μ C , are. placed at the vertices of an equilateral triangle of side. 10 cm. Find the work done by a person in pulling them apart to infinite separations.

MCQ: Work Done in Pulling Charges Apart

Three particles, each having a charge of \(10\,\mu C\), are placed at the vertices of an equilateral triangle with a side of \(10\,cm\). Find the work done by a person in pulling them apart to infinite separations.

• A) \(27\,J\)
• B) \(9\,J\)
• C) \(54\,J\)
• D) \(81\,J\)

Detailed Step-by-Step Solution

The work done in separating the charges is equal to the total electric potential energy of the system.

For a system of point charges, the potential energy \(U\) is given by: where \(k\) is Coulombs constant (\(\approx 9 \times 10^9 \, \text{Nm}^2/\text{C}^2\)), \(q_i\) and \(q_j\) are the charges, and \(r_{ij}\) is the distance between them.

For three identical charges placed at the vertices of an equilateral triangle, there are three pairs. Since all charges are \(q = 10\,\mu C = 10 \times 10^{-6}\,C\) and the distance between any two charges is \(r = 10\,cm = 0.1\,m\), the total potential energy is: \[ U = 3 \times \frac{k \, q^2}{r} \]

Substitute the given values: \[ U = 3 \times \frac{9 \times 10^9 \times (10 \times 10^{-6})^2}{0.1} \]

First, calculate the square of the charge: \[ (10 \times 10^{-6})^2 = 100 \times 10^{-12} = 1 \times 10^{-10} \]

Now substitute back: \[ U = 3 \times \frac{9 \times 10^9 \times 1 \times 10^{-10}}{0.1} = 3 \times \frac{0.9}{0.1} = 3 \times 9 = 27\,J \]

Therefore, the work done in pulling the charges apart to infinity is \(27\,J\).