Question

An electric dipole of length 10 cm having charges ± 6 × 10–3 c placed at 30° with respect to a uniform electric field experiences a torque of magnitude . find magnitude of electric field.

Answer 1

torque = PESin theta

6√3= (6*10^-3*0.1)*E*sin 30°

6√3=(6*10^-4)*E*1/2

6√3=3*10^-4*E

E=2√3*10^4

Answer 2

An electric dipole of a length 10cm having charges 6 × 10^-3C placed at 30° with respect to a uniform electric field experiences a torque of magnitude 6√3N.m, calculate the magnitude of electric field?

an electric dipole of length 10cm having charges ± 6 × 10^-3 C placed at 30° with respect to a uniform electric field.

so, dipole moment of dipole, P = qd

= 6 × 10^-3 C × (10cm)

= 6 × 10^-3 C × (10/100) m

= 6 × 10^-4 C.m

and angle between dipole , P and external electric field , E is , $\theta$ = 30°

we know, torque = P.Esin$\theta$

or, 6√3 = 6 × 10^-4 × E × sin30°

or, √3 × 10⁴ = E × (1/2)

or, E = 2√3 × 10⁴ N/C

hence, magnitude of electric field is 2√3 × 10⁴ N/C.