An electric dipole is placed at an angle of 30° with an electric field intensity of 2 × ${10}^5$ N/C. It experiences a torque equal of 4 Nm. The charge on the dipole, if the dipole length is 2 cm, is

Step 1: Given that:

Angle $\theta ={30}^0$

Electric field (E) = $2\times {10}^5$ N/C

Torque($\text{τ}$) = $4Nm$

Dipole length ($l$ ) = $2cm$ = $2\times {10}^{-2}m$

Step 2: Calculation of charge on the dipole:

We know that,

When an electric dipole is placed in a uniform electric field, it experiences a torque($\text{τ}$), given by:

$\text{τ}=pE\sin \theta$

Where,

$p=dipolemoment$

$=q\times l$

Therefore, torque will be:

$\text{τ}=qlE\sin \theta$

$q=\frac{\text{τ}}{lE\sin \theta }$

Putting all the values, we get

$q=\frac{4Nm}{2\times {10}^{-2}m\times 2\times {10}^{5}N{C}^{-1}\sin {30}^0}$

$q=\frac{4}{4\times {10}^3\times \frac{1}{2}}$

$q=2\times {10}^{-3}C$

$q=2mC$

Thus,

The charge on the dipole is $2mC$ .