Question

An electric dipole of length 2 cm is placed with its axis making an angle of 30° to a uniform electric field 10^5N/C.If it experiences a torque of 17.3Nm then potential energy of the dipole will be?

Answer 1

we know, torque of electric dipole is the cross product of dipole moment and electric field.

i.e., $\bf{\tau=P\times E}$

magnitude of torque, $|\tau|=|P||E|sin\theta$,

given, $|\tau|=17.3Nm,|E|=10^5N/C$ and $\theta=30^{\circ}$

so, 17.3N/m = |P| × 10^5 × sin30°

or, 3.46 × 10^-4 = |P|

so, magnitude of dipole moment, |P| = 3.46 × 10^-4 Cm

now, potential energy = -P.E

= -|P|.|E|cos30°

= -3.46 × 10^-4 × 10^5 × √3/2

= - 3 × 10¹

= -30 J

Answer 2

Answer:mark me brainliest if it helps

Explanation: