Question 843: Electrostats & Capacitance

Question

The work done in rotating a dipole placed parallel to the electric field through $$180^{\circ}$$ is W. So, the work done in rotating it through $$60^{\circ}$$ is $$\left(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{2}, \cos 180^{\circ}=-1\right)$$

Accepted Answer

$$\because$$ Work done in rotating dipole by $$	heta$$ angle from equilibrium,
$$\begin{aligned}
& W=p E[1-\cos 	heta] \quad 	ext{..... (i)}\
& \Rightarrow \quad W=p E\left[1-\cos 180^{\circ}ight] \
& \Rightarrow \quad W=p E[1+1]=2 p E \
& 	ext { or } \quad p E=\frac{W}{2} \quad 	ext{..... (ii)}\
& 	ext { and } \quad W^{\prime}=p E\left(1-\cos 60^{\circ}ight] \
& \Rightarrow \quad W^{\prime}=p E\left[1-\frac{1}{2}ight]=\frac{p E}{2} \
& \Rightarrow \quad W^{\prime}=\frac{p E}{2} \quad 	ext{..... (iii)}\
\end{aligned}$$
From Eq. (ii) put the value of $$p E$$ in Eq. (iii), we get
$$\Rightarrow \quad W^{\prime}=\frac{W}{2 	imes 2}=\frac{W}{4} \Rightarrow W^{\prime}=\frac{W}{4}$$

Answer

$$4 W$$

Answer

$$3 \mathrm{~W}$$

Answer

$$W / 2$$

Answer

$$W / 4$$

Electrostats & Capacitance

The work done in rotating a dipole placed parallel to the electric field through $180^{\circ} is W. So, the work done in rotating it through 60^{\circ} is \left(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{2}, \cos 180^{\circ}=-1\right)

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