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 movement of $$q_3$$ from C to D is given by $$\frac{q_3 K}{4\pi\epsilon_0}$$, the value of K is :

Given:

• AB=30 cm
• AC=AD=40 cm (circular path centered at A, so $$q_1$$​ is at center)

Thus distance of $$q_3$$ from $$q_1$$​ stays constant during motion:

So potential energy due to $$q_1$$​ does not change.

Only contribution from $$q_2$$ changes.

Distances from $$q_2$$:

At C:

$$r_{2c}=\sqrt{\ 30^2+40^2}=50cm$$

At D:

Since AD=40 and AB=30,

BD=10 cm

so

$$r_{2D}=10cm$$

Change in potential energy:

$$\Delta U=\frac{1}{4\pi\varepsilon_0}q_3q_2\left(\frac{1}{r_{2D}}-\frac{1}{r_{2C}}\right)$$

Substitute:

$$=\frac{1}{4\pi\varepsilon_0}q_3q_2\left(\frac{1}{10}-\frac{1}{50}\right)=\frac{1}{4\pi\varepsilon_0}q_3q_2\times\ \left(\frac{4}{50}\right)$$
convert 50 cm to m
$$=\frac{1}{4\pi\varepsilon_0}q_3q_2\times\ \left(\frac{4}{0.5}\right)$$
final answer is 
$$=\frac{8}{4\pi\varepsilon_0}q_3q_2$$
now comparing it with the given equation 
$$K=8q_2$$