Question

A small 15.0 g plastic ball is suspended by a 30.0 cm string in a uniform electric field of 
3000  N/C, as shown in the picture. If the ball is in equilibrium when the string makes a 45° angle with the vertical as indicated, what is the net charge on the ball?

Answer 1

Taking E $=$ Intensity of electric field

T$=$ Tension on string

M$=$ Mass of the ball

g$=$ Acceleration due to gravity

$q=$ Charge of the plastic ball

Since the ball is equilibrium $\sum$$F_y=0$ And $\sum\>F_x=0$

T sin $45=$ qE

T cos $45=$ mg

Tan $45°=\frac{qE}{mg}\implies\>q=\frac{mg\>tan45}{E}$

Where $m=\frac{15}{1000}=0.015kg$

$g=9.81$

$E=3000N/C$

$q=\frac{0.015×9.81\>tan45}{3000}$

$=4.905×10^{-5}C$