Question

PQRS is square of side 1 m. A charge of \(100 \, \mu C\) is placed at the centre of the square. Then the work done to take \(3 \, \mu C\) charge from the corner P to the corner R.

• A. \(9 \sqrt{2} \times 10^5 \, \text{J}\)
• B. \(4.5 \times 10^5 \, \text{J}\)
• C. Zero
• D. \(1.8 \times 10^5 \, \text{J}\)

Hint:

When a charge is moved in an electric field where the potential is the same at the start and end points, the work done is zero.

Answer 1

The Correct Option is C

The electric potential at any point due to a point charge is given by the formula: \[ V = \frac{kQ}{r} \] where \(V\) is the electric potential, \(k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2\) is Coulomb’s constant, \(Q\) is the charge, and \(r\) is the distance from the charge. In this case, the charge \(Q = 100 \, \mu C = 100 \times 10^{-6} \, \text{C}\) is placed at the center of the square PQRS. The side length of the square is \(1 \, \text{m}\). The diagonal of the square is the line joining the corner P and R, and it has a length \( \sqrt{2} \times 1 \, \text{m} = \sqrt{2} \, \text{m} \). Now, the work done in moving a charge \(q = 3 \, \mu C = 3 \times 10^{-6} \, \text{C}\) from point P to R is given by the equation: \[ W = q(V_R - V_P) \] where \(V_R\) and \(V_P\) are the electric potentials at points R and P respectively. Since the charge is at the center of the square and the distances from the center to all four corners are equal, the electric potential at all corners is the same. Therefore, the potential difference \(V_R - V_P = 0\). Hence, the work done is: \[ W = 3 \times 10^{-6} \times 0 = 0 \, \text{J} \] Thus, the work done to take the charge from P to R is zero.