Question

Two charges $ q_1 $ and $ q_2 $ are separated by a distance of 30 cm. A third charge $ q_3 $ initially at C as shown in the figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to the movement of $ q_3 $ from C to D is given by $ \frac{q_3 K}{4 \pi \epsilon_0} $, the value of $ K $ is:

• A. \( 8 q_2 \)
• B. \( 6 q_2 \)
• C. \( 8 q_1 \)
• D. \( 6 q_1 \)

Hint:

When calculating the change in potential energy due to the movement of a charge between two positions, consider the distances between the moving charge and the other charges involved.

Answer 1

The Correct Option is A

We are given that: The charge \( q_1 \) and \( q_2 \) are separated by 30 cm. A third charge \( q_3 \) is moved from point C to point D along a circular path of radius 40 cm. 
The change in potential energy is given by \( \frac{q_3 K}{4 \pi \epsilon_0} \). 
The potential energy of a system of charges is given by: \[ U = \frac{q_1 q_2}{4 \pi \epsilon_0 r} \] 
Where \( r \) is the distance between the charges. In this case, the distance changes as \( q_3 \) moves from C to D. 
The change in potential energy involves the interactions between \( q_3 \) and both \( q_1 \) and \( q_2 \), and after simplification, we find that \( K \) is proportional to \( q_2 \). 
Final Answer (1)  \( 8 q_2 \)

Answer 2

Given the expressions for \( u(C) \) and \( u(D) \): \[ u(C) = \left( \frac{kq_1}{40} + \frac{kq_2}{50} \right) q_3 \] \[ u(D) = \left( \frac{kq_1}{40} + \frac{kq_2}{10} \right) q_3 \] The change in potential energy \( \Delta u \) is: \[ \Delta u = |u(D) - u(C)| = kq_2 \left[ \frac{1}{10} - \frac{1}{50} \right] q_3 \] Simplifying: \[ \Delta u = kq_4 \times \frac{4q_3}{50} \] Now, equating to SI units: \[ \left( \frac{4q_2 q_3 \times 2}{4 \pi \epsilon_0} \right) \text{ SI unit} \leftrightarrow \frac{gk}{4 \pi \epsilon_0} \] Thus: \[ k = 8q_2 \] \[ \boxed{k = 8q_2} \]