Question 253: Electrostats & Capacitance

Question

A point charge q of mass m is suspended vertically by a string of length l. A point dipole of dipole
moment $$\overrightarrow p $$ is now brought towards q from infinity so that the charge moves away. The final
equilibrium position of the system including the direction of the dipole, the angles and distances is
shown in the figure below. If the work done in bringing the dipole to this position is N $$ 	imes $$ (mgh),
where g is the acceleration due to gravity, then the value of N is _________ . (Note that for three
coplanar forces keeping a point mass in equilibrium,
$${F \over {\sin 	heta }}$$
is the same for all forces, where F is any
one of the forces and $$	heta $$ is the angle between the other two forces)

Accepted Answer

U
i
= 0
$${U_f} = {{kqp} \over {{{\left( {2l\sin {\alpha  \over 2}} ight)}^2}}}$$ + mgh .... (i)
Now, from $$\Delta$$OAB,
$$\alpha$$ + 90 $$-$$ $$	heta$$ + 90 $$-$$ $$	heta$$ = 180
$$\Rightarrow$$ $$\alpha$$ = 2$$	heta$$
From $$\Delta$$ABC, h = 2l sin$$\left( {{\alpha  \over 2}} ight)$$ sin$$	heta$$
h = 2l sin$$\left( {{\alpha  \over 2}} ight)$$ sin$$\left( {{\alpha  \over 2}} ight)$$ $$\Rightarrow$$ h = 2l sin
2
$$\left( {{\alpha  \over 2}} ight)$$
Now, charge is in equilibrium at point B.
So, using sine rule,
$${{mg} \over {\sin \left( {90 + {\alpha  \over 2}} ight)}} = {{qE} \over {\sin 1(80 - 2	heta )}}$$
$$ \Rightarrow {{mg} \over {\cos {\alpha  \over 2}}} = {{qE} \over {\sin 2	heta }}$$
$$ \Rightarrow {{mg} \over {\cos {\alpha  \over 2}}} = {{qE} \over {\sin \alpha }} \Rightarrow {{mg} \over {\cos {\alpha  \over 2}}} = {{qE} \over {2\sin {\alpha  \over 2}\cos {\alpha  \over 2}}}$$
$$ \Rightarrow qE = mg2\sin \left( {{\alpha  \over 2}} ight)$$
$$ \Rightarrow {{q2kp} \over {{{\left( {2l\sin {\alpha  \over 2}} ight)}^3}}} = mg2\sin \left( {{\alpha  \over 2}} ight)$$
$$ \Rightarrow {{kpq} \over {{{\left( {2l\sin {\alpha  \over 2}} ight)}^2}}} = mg\sin \left( {{\alpha  \over 2}} ight) 	imes \left( {2l\sin {\alpha  \over 2}} ight)$$
$$ \Rightarrow {{kpq} \over {{{\left( {2l\sin {\alpha  \over 2}} ight)}^2}}} = mgh$$
Substituting this in Eq. (i), we get
$${U_f} = mgh + {{kpq} \over {{{\left( {2l\sin {\alpha  \over 2}} ight)}^2}}} \Rightarrow {U_f} = 2mgh$$
W = $$\Delta$$U = 2mgh = Nmgh
$$ 	herefore $$ N = 2

Electrostats & Capacitance

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