Question

An electric dipole, when held at 30 degree with respect to a uniform electric field of 10^4 N/C, experiences a torque of 9×10^-26Nm. Calculate the moment of the dipole

Answer 1

We know that

Dipole moment is denoted by ρ

Also that

torque=τ

and we know that

τ=ρΧΕ

or we can also write it as

τ=ρΕsinθ

as it is a cross (vector product)

Hence substituting the values

$9\times 10^{-26} =\rho \times 10^{4} sin(30)\\\\9\times 10^{-26}=\rho \times 10^{4}\times \frac{1}{2}\\\\ 9\times 10^{-26}=\rho\times5\times 10^{3}\\\rho=\frac{9\times 10^{-26}}{5\times 10^{3}}\\\rho =1.8\times10^{-29}Cm$

Answer 2

Given,

E = $10^{4}$ N/C

τ = 9×$10^{-26}$ Nm

θ = 30°

(all the quantities are in SI unit, so no conversion is required.)

To find,

The dipole moment of the given electric dipole.

Solution,

We know that torque (τ) is given by,

τ = PEsinθ (where P is the dipole moment of the dipole.)

Here, if we put the values of τ, E, and θ in the equation, we will get

⇒ 9×$10^{-26}$ = P × $10^{4}$ × sin30°

⇒ 9×$10^{-26}$ = P × $10^{4}$ × 0.5

⇒ P = $\frac{ 9X10^{-26} }{10^{4}X0.5 }$

Solving the above equation we get,

P = 18×$10^{-30}$ Cm

(This is the final answer as all the physical quantities were given in SI units and the final answer is also in SI units.)