Question

An electric dipole held at an angle of 300 with respect to a Uniform electric field of 4 x 104 N/C experiences a torque of 1.6 x10-25 Nm calculate its dipole Moment.

Answer 1

given,
x=300 degree
electric field E=4x10^4N/C
torque =1.6 x10^-25Nm
torque = px E Sinx where x=300 degree
dipole moment p=torque /Esinx
p=1.6x10^-25/4x10^4Sin300 [Sin300=-0.867]
p=1.6x10^-25//4.5X10^4x(-0.867)
p=1.6x10^-25/3.90x10^4
p=-0.410x10^-29Nm
I HOPE THIS WILL HELP YOU.....