Question

An electric dipole consists of $1.0g$spheres charged to $\{2.0nC$at the ends of a $10cm$-long massless rod. The dipole rotates on a frictionless pivot at its center. The dipole is held perpendicular to a uniform electric field with field strength $1000V/m$, then released. What is the dipole’s angular velocity at the instant it is aligned with the electric field?

Short answer

The dipole’s angular velocity is$0.28rad/s$.

Step-by-step solution

Given Information

We have given a dipole consists of$1.0g$spheres, charge$\pm 2.0nC$, length of massless rod$10cm$and a uniform electric field with field strength$1000V/m$.

Law of conservation of energy

By the law of conservation of energy, the entire energy that the dipole has at the moment it is aligned perpendicular to the electric field will be converted to kinetic energy, which in this case is rotational. Consider the formula for the energy of a rotating system

$E_k=\frac{I{w}^2}{2}$,

Where $I$is the moment of inertia. For our case, the moment of inertia will be

$I=2·m{(d/2)}^2=\frac{md2}{2}$,

This is because the moments come from the two spheres, each of which connected by a massless rod, keeping them at $d/2$distance from the axis of rotation. Found that the kinetic energy is $E_k=\frac{m{d}^2{w}^2}{4}$

Therefore, we can express the angular velocity as

$w=\frac{2}{d}\sqrt{\frac{E_k}{rn}}$

Explanation

Now, as we mentioned in the introduction, finding the kinetic energy in terms of the electric potential energy.

$E_k=E_p=-pE$

Where $p=qd$is the magnitude of the dipole moment and $E$is the magnitude of the electric field.

Let us disregard the minus sign for simplicity, and have

$E_k=qdE$.

Substituting this result in our result for the angular velocity we will have

$w=\frac{2}{d}\sqrt{\frac{qdE}{m}}=2\sqrt{\frac{qE}{md}}$

As the expression can't be simplified further. We can now calculate and have

$w=2\sqrt{\frac{2·{10}^{-9}·{10}^3}{1·{10}^{-3}·0.1}}\approx 0.28rad/s$

$w=\frac{2}{d}\sqrt{\frac{qdE}{m}}=2\sqrt{\frac{qE}{md}}$