Question

An electric dipole of length 2 cm is placed with its axis at an angle of 30° to a uniform electric field of \[{{10}^{5}}\] N/C. It experiences a torque of 10√3 N-m. The potential energy of the dipole is and the magnitude of charge on the dipole?

Answer 1

Hint: a dipole is a combination of two charges of equal magnitude but opposite in polarity separated by some distance. When a dipole is placed parallel to an electric field the net force on it is zero and there is no torque. But when it is placed at some angle with the field then it experiences net torque.

Step by step answer:
We know torque on a dipole in an electric field making some angle, \[\zeta \] with it is given by the formula,
\[\tau =pE\sin \zeta \]
Dipole moment, p =\[q\times 2r\], here q is the magnitude of each charge and r is separation between them.
Given the value of r is 2cm= 0.02 m
Given the value of the filed and the angle that the dipole makes with it, substituting these values we get,
\[
  10\sqrt{3}=q\times 2r\times {{10}^{5}}\sin 30 \\
  \Rightarrow 10\sqrt{3}=q\times 0.06\times {{10}^{5}} \\
  \Rightarrow q=2.88\times {{10}^{-3}}C \\
\]
\[\Rightarrow q=2.88\times {{10}^{-3}}C\]
Thus, the charges are \[\pm 2.88\times {{10}^{-3}}C\] and they are separated by 3 cm distance.
Now we need to find the value of potential energy, it is given by the formula,
\[
 U=-pE\cos \zeta \\
 \Rightarrow U=-q\times 2r\times E\times \cos 30 \\
\Rightarrow U=-2.88\times {{10}^{-3}}\times {{10}^{5}}\times 2\times 0.03\times \Rightarrow \dfrac{\sqrt{3}}{2} \\
\Rightarrow U=-14.96J \\
\]
\[\therefore U=-14.96J\]

Hence potential energy is obtained as 14.96J.

Note: Potential energy of a dipole comes out to be negative because the positive charge is placed at a location with a higher potential than the negative charge. Also, negative value tells us that the configuration is stable. If we draw the potential energy vs distance graph, we see the curve is bounded.