Two charges q₁ and q₂ are placed 30 cm apart as shown. A third charge q₃ is moved along the circle of radius 40 cm from C to D. The change in the potential energy of the system is $\frac{q_{3}K}{4\pi {\in }_0}$ Find K?

Given two charges q1 and q2 are placed 30 cm apart as shown in the figure. A third charge q3 moved from C to D.
Radius  = 40 cm$k=q_2\left(10-2\right)=8q_2$
Given the change in potential energy of the system is $\frac{q_{3}k}{4\pi {\in }_0}$. We have to find the value of k. 
We solve it as follows:
The change in potential energy of system is
$U_D-U_C$ as discussed under: When charge $q_3$
is at C, then its potential energy is 
$U_C=\frac{1}{4\pi {\in }_0}\left[\frac{q_1{q}_3}{0.4}+\frac{q_2{q}_3}{0.5}\right]$
Where charge q₃ is at D, then U_D is
$U_D=\frac{1}{4\pi {\in }_0}\left[\frac{q_1{q}_3}{0.4}-\frac{q_2{q}_3}{0.1}\right]$
Hence, change in potential energy is ΔU
$U_D=\frac{1}{4\pi {\in }_0}\left[\frac{q_2{q}_3}{0.1}-\frac{q_2{q}_3}{0.5}\right]\dots \dots .\left(i\right)$
But given, $\mathrm{\Delta U}=\frac{q_{3}k}{4\pi {ϵ}_0}\dots \dots \left(ii\right)$
On equating (i) and (ii), we get
$\frac{q_{3}k}{4\pi {\in }_0}=\frac{1}{4\pi {\in }_0}\left[\frac{q_2{q}_3}{0.1}-\frac{q_2{q}_3}{0.5}\right]$
We get,
$k=q_2\left(10-2\right)=8q_2$