Question 11: Electrostatics & Capacitance

Question

Charges $$+q$$ and $$-q$$ are placed at points $$A$$ and $$B$$ respectively which are a distance $$2 L$$ apart, $$C$$ is the mid-point between $$A$$ and $$B$$. The work done in moving a charge $$+Q$$ along the semicircle $$C R D$$ is

Accepted Answer

In Ist case, when charge +Q is situated at C
Electric potential energy of system
$$U_1=\frac{1}{4 \pi \varepsilon_0} \frac{q(-q)}{2 L}+\frac{1}{4 \pi \varepsilon_0} \frac{(-q) Q}{L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}$$
In Ind case, when charge $$+Q$$ is moved from $$C$$ to $$D$$
Electric potential energy of system in this
case
$$U_2=\frac{1}{4 \pi \varepsilon_0} \frac{q(-q)}{2 L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{3 L}+\frac{1}{4 \pi \varepsilon_0} \frac{(-q)(Q)}{L}$$
Work done $$(\Delta U)=U_2-U_1$$
$$\begin{aligned}
= & \left(-\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2 L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{3 L}-\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}ight) \
& -\left(-\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2 L}-\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}ight) \
= & \frac{q Q}{4 \pi \varepsilon_0}\left[\frac{1}{3 L}-\frac{1}{L}ight]=\frac{q Q}{4 \pi \varepsilon_0} \frac{(1-3)}{3 L} \
= & -\frac{2 q Q}{12 \pi \varepsilon_0 L}=-\frac{q Q}{6 \pi \varepsilon_0 L}
\end{aligned}$$

Answer

$$\frac{q Q}{4 \pi \varepsilon_0 L}$$

Answer

$$\frac{q Q}{2 \pi \varepsilon_0 L}$$

Answer

$$\frac{q Q}{6 \pi \varepsilon_0 L}$$

Answer

$$\frac{-q Q}{6 \pi \varepsilon_0 L}$$

Electrostatics & Capacitance

Charges $+q and -q are placed at points A and B respectively which are a distance 2 L apart, C is the mid-point between A and B. The work done in moving a charge +Q along the semicircle C R D$ is

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