Question

Two charges q₁ and q₂ are placed 30 cm apart, as shown in the figure. A third charge q₃ is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is\(\bigg(\frac{q_3}{4\pi\in_0}\bigg)k\),where k is:

• A. 8q₂
• B. 8q₁
• C. 4q₂
• D. 4q₁

Answer 1

The Correct Option is A

 To solve this problem, we need to determine the change in potential energy as charge \( q_3 \) is moved from point C to point D along the arc. The potential energy change is given by:

\(U = \bigg(\frac{q_3}{4\pi \varepsilon_0}\bigg)k\)

We need to find the value of \( k \).

The potential energy at any point is given by:

\(U = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q_1 q_3}{r_{13}} + \frac{1}{4\pi \varepsilon_0} \cdot \frac{q_2 q_3}{r_{23}}\)

where \( r_{13} \) is the distance from \( q_1 \) to \( q_3 \) and \( r_{23} \) is the distance from \( q_2 \) to \( q_3 \).

Step 1: Calculate \( r_{13} \) and \( r_{23} \).

• At point C, \( r_{13} = 40 \text{ cm} \), \( r_{23} = 50 \text{ cm} \) (using the Pythagorean theorem in triangle \( \bigtriangleup CAB \)).
• At point D, \( r_{13} = 50 \text{ cm} \), \( r_{23} = 30 \text{ cm} \).

Step 2: Calculate the initial and final potential energy.

• Initial potential energy at C:
• Final potential energy at D:

Step 3: Determine the change in potential energy.

\(\Delta U = U_D - U_C = \frac{q_3}{4\pi \varepsilon_0} \left(\frac{q_1}{50} - \frac{q_1}{40} + \frac{q_2}{30} - \frac{q_2}{50}\right)\)

Simplifying,

\(\Delta U = \frac{q_3}{4\pi \varepsilon_0} \left(-\frac{q_1}{200} + \frac{16q_2}{300}\right) = \frac{q_3}{4\pi \varepsilon_0} \cdot 8q_2\)

Thus,

\(k = 8q_2\)

The correct answer is 8q₂.