Question

A 3.0 g aluminum foil ball with a charge of 3.0×10−9C hangs freely from a 1.4 m-long thread. What angle with the vertical the equilibrium position of the string makes?

Answer 1

Answer:

0 degree

Explanation:

Mass of aluminium foil=m=3 g=$\frac{3}{1000}=0.003 kg$

1 kg=1000 g

$q=3\times 10^{-9} C$

Length of thread=l=1.4 m

We have to find the angle made by the string with the vertical equilibrium position.

The ball hang freely therefore, there is no field present.

The angle made by equilibrium position of the string with the vertical is zero because it is influenced under the acceleration due to gravity of earth and equilibrium position  lies along the vertical axis.

Hence, the angle made by the equilibrium position of the string =0 degree