Question

an electric dipole is placed at an angle 60° with an electric flied of intensity 1000000 N/C. it experience a torque equal to 8√3 NM. calculate charge on dipole, if dipole length is 2cm

Answer 1

Torque on dipole, T = P x E x Sin( ¥ )
Where
P = Dipole Moment = 2qr = q x ( 2 r )
Here q = Charge on dipole ,
2r = length of dipole = 2cm = 0.02m
E = Electric Field intensity = 1000000 N/C
¥ = angle between P and E = 60 degree

Now
T = P x E x Sin ( ¥ )
= { q x ( 2r ) } x E x Sin ( ¥ )
= q x 2r x E x Sin ( ¥ )
q = T / 2r x E x Sin ( ¥ )
= 8 ( Root 3 ) / 0.02 x 1000000 x ( ( Root 3)/2)
= [8 ( Root 3 ) x 100 x 2 ] / 2 x 1000000 x ( Root 3 )
= 8 / 10000
= 8 x 10^ - 4 C
= 8 x 10^ - 4 C x { 100 / 100 }
= { 8 x 100} x { 10 ^ - 4 / 100 } C
= 800 x 10 ^ - 6 C
= 800 microCoulomb

Answer 2

The formula for torque when an electric dipole I'd placed in electric field is

Torque = PE Sin tetha

P = 2r(q) [ Dipole moment]

E = 1000000N/C

8√3 = 2 × 10^-2 (q) 10^6 Sin 60°

q = 8√3 ×2 / 2 × 10^4√3

q = 8 × 10-4 C

:)