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Question

(C) The initial potential energy of the system is $U_i = 0$.The final potential energy of the system is $U_f = \frac{k q p}{(2l \sin(\alpha/2))^2} + m

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The initial potential energy of the system is $U_i = 0$.The final potential energy of the system is $U_f = \frac{k q p}{(2l \sin(\alpha/2))^2} + mgh$,where $k = \frac{1}{4\pi\epsilon_0}$.From the geometry of the triangle,the distance $r = 2l \sin(\alpha/2)$.For the charge $q$ in equilibrium,the forces acting are tension $T$,gravity $mg$,and the electric force $F_e = qE$. The electric field of a dipole at distance $r$ is $E = \frac{kp}{r^3} \sqrt{1 + 3\cos^2\phi}$. Given the geometry,the force $F_e = \frac{kqp}{r^3} 	imes 2$ (for the specific orientation).Using the Lami's theorem or the sine rule for equilibrium:$\frac{mg}{\sin(90^\circ + \alpha/2)} = \frac{qE}{\sin(180^\circ - 2	heta)}$.Solving the equilibrium condition leads to $F_e = mg \cdot 2 \sin(\alpha/2)$.Substituting $F_e = \frac{kqp}{r^2} \cdot \frac{2}{r} = \frac{2kqp}{r^3}$,we find that the potential energy $U_f = \frac{kqp}{r^2} + mgh = mgh + mgh = 2mgh$.Since $W = \Delta U = U_f - U_i = 2mgh - 0 = 2mgh$,we have $N = 2$.

List-$I$	List-$II$
$(P)$ Two dipoles pointing in $+\hat{j}$ direction at $x = -r$ and $x = +r$	$(1) \ \vec{E}=0$
$(Q)$ Two dipoles pointing in $+\hat{j}$ and $-\hat{j}$ direction at $x = -r$ and $x = +r$ respectively	$(2) \ \vec{E}=-\frac{p}{2 \pi \epsilon_0 r^3} \hat{j}$
$(R)$ Two dipoles pointing in $+\hat{j}$ and $+\hat{i}$ direction at $x = -r$ and $x = +r$ respectively	$(3) \ \vec{E}=-\frac{p}{4 \pi \epsilon_0 r^3}(\hat{i}-\hat{j})$
$(S)$ Two dipoles pointing in $+\hat{i}$ direction at $x = -r$ and $x = +r$	$(4) \ \vec{E}=\frac{p}{4 \pi \epsilon_0 r^3}(2\hat{i}-\hat{j})$
	$(5) \ \vec{E}=\frac{p}{\pi \epsilon_0 r^3} \hat{i}$

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