Question

A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 10^3 N/C. If the ball is in equilibrium when the string makes a 30 o angle with the vertical, what is the net charge on the ball? Q =__________

Answer 1

Answer:

The net charge is $67.89 \mu C$

Solution:

As per the question:

Mass of the plastic bag, m = 12.0 g = $12.0\times 10^{-3}\ kg$

Magnitude of electric field, E = $10^{3}\ N/C$

Angle made by the string, $\theta = 30^{\circ}$

Now,

To calculate the net charge, Q on the ball:

Vertical component of the tension in the string, $T = Tcos\theta$

Horizontal component of the tension in the string, $T = Tsin\theta$

Now,

Balancing the forces in the x-direction:

$Tsin\theta = QE$

$Q = {Tsin\theta}{E}$                             (1)

Balancing the forces in the y-direction:

$Tcos\theta = mg$

where

g = acceleration due to gravity = $9.8\ m/s^{2}$

Thus

$T = \frac{mg}{cos\theta }$

$T = \frac{12.0\times 10^{-3}\times 9.8}{cos30^{\circ}} = 0.1357\ N$

Use T = 0.1357 N in eqn (1):

$Q = {0.1357\times sin30^{\circ}}{10^{3}} = 6.789\times 10^{- 5}\ C$

$Q = 67.89\times 10^{- 5}\ C = 67.89\mu C$