Question

A small 2.00 g plastic ball is suspended by a 20.0 cm long string in a uniform electric field. If the ball is in equilibrium when the string makes a 15.0 degree angle with the vertical, what is the net charge on the ball?

Answer 1

Let’s draw a free-body diagram:

Here, $W=mg$ is the force of gravity (or weight of the plastic ball), $F_e=qE$ is the electric force that acts on the plastic ball, $T$ is the force of tension in the string, $T_x$,$T_y$ is the projections of the force of tension on axis $x$ and $y$, respectively, $\theta$ is the angle of inclination of the string to the vertical, $q$ is the charge of the plastic ball and $E$ is the intensity of the electric field.

Let’s write the conditions of the equilibrium for the plastic ball:

$$\sum F_x=0, \sum F_y=0.$$

Let’s consider the forces that act on the plastic ball in the horizontal $x$- and vertical $y$-direction:

$$\sum F_x=F_e-Tsin\theta=qE-Tsin\theta=0, (1)$$ $$\sum F_y=Tcos\theta-mg=0. (2)$$

We can express the force $T$ from the second equation:

$$T=\dfrac{mg}{cos\theta}.$$

Then, we can substitute it into the first equation and get:

$$qE-mgtan\theta=0,$$ $$qE=mgtan\theta.$$

From the last equation we can find the magnitude of the net charge on the ball:

$$q=\dfrac{mgtan\theta}{E}.$$

Unfortunatelly, we don't know the intensity of the electric field. Let's suppose, for example, that $E=1000\ \dfrac{N}{C}.$ Then, we can calculate the magnitude of the net charge on the ball:

$$q=\dfrac{0.002\ kg\cdot 9.8\ \dfrac{m}{s^2}\cdot tan15^{\circ}}{1000\ \dfrac{N}{C}}=5.25\cdot 10^{-6}\ C$$

Answer:

$q=5.25\cdot 10^{-6}\ C$