Two charges $\mathrm{q}_1$ and $\mathrm{q}_2$ 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 $\frac{\mathrm{q}_3 \mathrm{~K}}{4 \pi \epsilon_0}$, the value of K is :

Question

Aicharya · Accepted Answer

Potential at C
$\mathrm{V}_{\mathrm{C}}=\frac{\mathrm{kq}_1}{0.4}+\frac{\mathrm{kq}_2}{0.5}$
Potential at D
$\mathrm{V}_{\mathrm{D}}=\frac{\mathrm{kq}_1}{0.4}+\frac{\mathrm{kq}_2}{0.1}$
$\Delta \mathrm{U}=\left(\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{c}}\right)\left(\mathrm{q}_3\right)=\left(\frac{\mathrm{kq}_2}{0.1}-\frac{\mathrm{kq}_2}{0.5}\right)\left(\mathrm{q}_3\right)$
$\Delta \mathrm{U}=8 \mathrm{kq}_2 \mathrm{q}_3=\frac{8 \mathrm{q}_2 \mathrm{q}_3}{4 \pi \varepsilon_0}$

Two charges $\mathrm{q}_1$ and $\mathrm{q}_2$ are separated by a distance of 30 cm. A third charge $q_3$…

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