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 $q_3 k/4\pi\varepsilon_0$, the value of K is ($\varepsilon_0$ = permittivity of free space)

The initial potential energy of the system when charge $q_3$ is at point C is given by:

$U_C = \dfrac{1}{4\pi\varepsilon_0} \left( \dfrac{q_1 q_3}{AC} + \dfrac{q_2 q_3}{BC} \right)$

From the given figure, $AC = 40\text{ cm} = 0.4\text{ m}$ and $AB = 30\text{ cm} = 0.3\text{ m}$.
In the right-angled triangle ABC, the distance BC is:
$BC = \sqrt{AB^2 + AC^2} = \sqrt{(0.3)^2 + (0.4)^2} = 0.5\text{ m}$

So, $U_C = \dfrac{q_3}{4\pi\varepsilon_0} \left( \dfrac{q_1}{0.4} + \dfrac{q_2}{0.5} \right)$

When $q_3$ is moved to point D along the circular path centered at A, the distance AD is equal to the radius of the path:
$AD = AC = 40\text{ cm} = 0.4\text{ m}$

The distance BD is:
$BD = AD - AB = 0.4\text{ m} - 0.3\text{ m} = 0.1\text{ m}$

The final potential energy of the system when $q_3$ is at point D is:

$U_D = \dfrac{1}{4\pi\varepsilon_0} \left( \dfrac{q_1 q_3}{AD} + \dfrac{q_2 q_3}{BD} \right) = \dfrac{q_3}{4\pi\varepsilon_0} \left( \dfrac{q_1}{0.4} + \dfrac{q_2}{0.1} \right)$

The difference in potential energy is:

$\Delta U = U_D - U_C = \dfrac{q_3}{4\pi\varepsilon_0} \left( \dfrac{q_1}{0.4} + \dfrac{q_2}{0.1} - \dfrac{q_1}{0.4} - \dfrac{q_2}{0.5} \right)$

$\Delta U = \dfrac{q_3}{4\pi\varepsilon_0} \left( \dfrac{q_2}{0.1} - \dfrac{q_2}{0.5} \right)$

$\Delta U = \dfrac{q_3 q_2}{4\pi\varepsilon_0} \left( 10 - 2 \right) = \dfrac{q_3 (8q_2)}{4\pi\varepsilon_0}$

We are given that the difference in potential energy is $\dfrac{q_3 k}{4\pi\varepsilon_0}$.

Comparing the two expressions, we get:
$k = 8q_2$

Answer: $8q_2$