Question 623: Electrostats & Capacitance

Question

An electric dipole has a fixed dipole moment $$\overrightarrow p $$, which makes angle $$	heta$$ with respect to x-axis. When
subjected to an electric field  $$\mathop {{E_1}}\limits^ 	o   = E\widehat i$$ , it experiences a torque $$\overrightarrow {{T_1}}  = 	au \widehat k$$ . When subjected to another electric
field $$\mathop {{E_2}}\limits^ 	o   = \sqrt 3 {E_1}\widehat j$$  it experiences a torque $$\mathop {{T_2}}\limits^ 	o   = \mathop { - {T_1}}\limits^ 	o  $$ . The angle $$	heta$$ is:

Accepted Answer

Torque experienced by the dipole in an
electric field,
$$T $$ = pE sin$$	heta $$
$$\overrightarrow T   = \overrightarrow p  	imes \overrightarrow E $$
$$\overrightarrow p  = p\cos 	heta \widehat i + p\sin 	heta \widehat j$$
$$\mathop {{E_1}}\limits^ 	o   = E\widehat i$$
$$\overrightarrow {{T _1}}  = \overrightarrow P  	imes {\overrightarrow E _1}$$
= ($$p\cos 	heta \widehat i + p\sin 	heta \widehat j$$) $$ 	imes $$ $$E\left( {\widehat i} ight)$$
= pE sin$$	heta $$$$\left( { - \widehat k} ight)$$
$$\mathop {{E_2}}\limits^ 	o   = \sqrt 3 {E_1}\widehat j$$
$$\overrightarrow {{T _2}}  = $$($$p\cos 	heta \widehat i + p\sin 	heta \widehat j$$) $$ 	imes $$ $$\sqrt 3 {E_1}\widehat j$$
= $$\sqrt 3 pE\cos 	heta \left( {\widehat k} ight)$$
Now given, $$\overrightarrow {{T _2}}$$ = $$-\overrightarrow {{T _1}}$$
$$ \Rightarrow $$ $$\sqrt 3 pE\cos 	heta \left( {\widehat k} ight)$$ = -pE sin$$	heta $$$$\left( { - \widehat k} ight)$$
$$ \Rightarrow $$ $$	an 	heta  = \sqrt 3 $$
$$ \Rightarrow $$ $$	heta $$ = 60
o

Answer

90o

Answer

45o

Answer

30o

Answer

60o

Electrostats & Capacitance

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