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

From the figure, $AC = 40\text{ cm} = 0.4\text{ m}$ and $AB = 30\text{ cm} = 0.3\text{ m}$.

Triangle ABC is right-angled, so we use Pythagoras theorem to find $BC$.

$BC = \sqrt{AB^2 + AC^2} = \sqrt{(0.3)^2 + (0.4)^2} = 0.5\text{ m}$

So, the potential energy at C becomes:

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

Now $q_3$ is moved from C to D along a circular path of radius $40\text{ cm}$ with centre at A. So the distance from A remains the same.

$AD = AC = 40\text{ cm} = 0.4\text{ m}$

From the figure, points A, B, and D lie on the same straight line. So the distance $BD$ is:

$BD = AD - AB = 0.4\text{ m} - 0.3\text{ m} = 0.1\text{ m}$

The potential energy of charge $q_3$ 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 change in potential energy is:

$\Delta U = U_D - U_C$

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

The terms containing $q_1$ cancel because the distance of $q_3$ from $q_1$ remains $0.4\text{ m}$ at both C and D.

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

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

$\Delta U = \dfrac{q_3 (8q_2)}{4\pi\varepsilon_0}$

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

Comparing both expressions:

$k = 8q_2$

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