Question

an electric dipole of length 2cm is placed with its axis making an angle of 30`c to a uniform electric field of 100000 N/c if it experiences a torque of 10rt3 Nm calculate magnitude of charges on dipole

Answer 1

Torque = PEsin(thetha)
P = q(2* 10^-2) = 0.02 q( it is the electric dipole moment )
Given E= 100000
sin thetha = sin 30 = 0.5
hence Torque = 0.02* 100000* 0.5* q= 10000
q = 0.1 C

Answer 2

Magnitude of charges on dipole, q = 17.3 mC

Given:  

Length of “electric dipole”, d = 2 cm

Axis at an angle, $\theta$ = 30 degree  

Electric field, E = $10^5$ N/C

Torque = $10 \sqrt{3}$

Solution:  

Then, the torque is given by torque = $q d E \sin \theta$

Thereby,

$q=\frac{\text {Torque}}{d E \sin \theta}$

Substituting the values, we get,

$q=\frac{10 \sqrt{3}}{0.02 \times 10^{5} \times \sin 30}$

$=\frac{10 \sqrt{3}}{0.02 \times 10^{5} \times 0.5}$

$=0.01732 \ C$

$q = 17.3 \  mC$