Question

An electric dipole consists of two opposite charges each 0.05microcoulomb separated by 30mm. The dipole is placed in a uniform external electric field of 1000000NC-1.The maximum torque exerted by the field on the dipole is

Answer 1

Torque , T = P × E = (P)(E)(Sin@) where ,

P = Dipole moment ,

E = Electric field = 1000000 NC^-1 = 10^6 NC^-1

@ = angle between P and E = 90° as torque is maximum...

P = (2q ) × r where

q = charge = 0.05 uC = 5 × 10^-2 × 10^-6 C

= 5 × 10 ^-8 C

r = seperation between charges q = 30 mm

= 30 × 10 ^-3 m = 3 × 10^-2 m

P = (2q ) × r = 2 × 5 × 10^-8 × 3 × 10^-2

= 30 × 10^-10 = 3 x 10^-9 C-m

P = 3 x 10^-9 C-m

Torque ,

T = P × E × Sin@

= 3 × 10^-9 × 10^-6 × Sin90°

= 3 × 10^-15 × 1 ( since sin90 = 1 )

= 3 × 10^-15 Nm

Answer 2

Answer:

The maximum torque exerted by the field on the dipole is $1.5 \times 10 ^{-3} Nm$.

Explanation:

Given the charge of an electric dipole,

$q = 0.05 \mu C = 5 \times 10^{-2} \times 10^{-6}= 5 \times 10 ^{-8} C$

The separation between charges,  $d = 30 mm= 30 \times 10 ^{-3} m = 3 \times10^{-2} m$

Uniform external electric field,  $E = 1000000 NC^{-1} = 10^6 NC^{-1}$

A dipole moment, p is given by the product of the magnitude of the charge and the distance between the charges in a system, i.e., $p=qd$

Torque acting on an electric dipole placed in an uniform electric field is given by

$\vec \tau=\vec p\times \vec E \implies |\vec \tau|= p Esin\theta$

For maximum torque, $\theta=90^o$  

Substituting the given values,  

$\tau= p Esin\theta$  

$= 5 \times 10 ^{-8} \times 3 \times10^{-2} \times 10^{6} \times sin90^{o}$

$= 5 \times 10 ^{-4} \times 3$

$= 15 \times 10 ^{-4} = 1.5 \times 10 ^{-3} Nm$

Therefore, the maximum torque exerted by the field on the dipole is $1.5 \times 10 ^{-3} Nm$.