Question

Two charges \(q_1\) and \(q_2\) are separated by a distance of 30cm. A third charge \(q_3\) initially at point C is shown in figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to movement of \(q_3\) from C to D is \(q_3k/4πε_0\), the value of K is (\(ε_0\) = permittivity of free space)

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

Hint:

Only q2 changes its distance to q3; q1 stays at a fixed 40 cm because the path is a circle about q1.

Answer 1

The Correct Option is A

Step 1: Potential at C and D
$V_C = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{0.4}+\frac{q_2}{0.5}\right)$, $V_D = \frac1{4\pi\varepsilon_0}\left(\frac{q_1}{0.4}+\frac{q_2}{0.1}\right)$.

Step 2: Difference
$V_D - V_C = \frac{q_2}{4\pi\varepsilon_0}(10 - 2) = \frac{8q_2}{4\pi\varepsilon_0}$.

Step 3: Energy
Multiply by $q_3$: $K = 8q_2$. Option (A).

Final Answer:
8 q2. \[ \boxed{\text{(A)}\ K=8q_2} \]