Question

An electric dipole placed in a uniform electric field of magnitude 2 N/C, experiences a maximum torque of magnitude 5 x 10-3 Nm. If the charges of the dipole have a magnitude of 2 mC, then the length of the dipole is​

Answer 1

Given:

An electric dipole placed in a uniform electric field of magnitude 2 N/C, experiences a maximum torque of magnitude 5 x 10-3 Nm. Magnitude of charge is 2 mC.

To find:

Length of dipole ?

Calculation:

• We know that torque experienced is maximum when angle between field vector and dipole is 90°.

$\tau = P \times E \times \sin( {90}^{ \circ} )$

• Putting values in SI UNITS:

$\implies 5 \times {10}^{ - 3} = (q \times l) \times 2$

$\implies 5 \times {10}^{ - 3} = (2 \times {10}^{ - 3} \times l) \times 2$

$\implies l = 1.25 \: metres$

So, length of dipole is 1.25 metres.

Answer 2

Given:-

An electric dipole place in a uniform electric field of magnitude 2N/C , which is experiences a maximum torque of magnitude 5*10^-3 Nm. And the charges of the dipole are 2mC.

To Find:-

The Length of the dipole

Solution:-

As we know the formula of the torque(τ) i.e.

                τ = P*E*Sinθ , And we know that   P = q*l

• Where 'P' is the electric dipole , 'E' is the electric field , 'q' is the charge and 'l' is the length b/w the two charges.

• So, the final equation becomes =>   τ = (q*l)*E*Sinθ
• Since here it is given the maximum torque it means that the angle'θ' should be of 90°.
• All values should be written as per the S.I unit of system so convert 2mC into C that will be 2*10^-3C.

Now put all the values we get;

(5*10^-3 Nm) =(2*10^-3 C *l)*2N/C*Sin90°  

=> l = 5*10^-3 / 2*1*2*10^-3            {∴ Sin90°=1}

=> l=  1.25m

The Length(l) of the dipole is 1.25m.