Question

A uniform horizontal electric field of 1.8 × 105 N/C causes a ball that is suspended from a light string to hang at an angle of 23° from the vertical. If the mass of the ball is 5.0 grams, what is the magnitude of its charge?

Answer 1

`1.15669* 10^(-7)\ C`

Step-by-step explanation:

`\theta` = Angle with which the electric field is hung = 23°

m = Mass of ball = 5 g

E = Electric field =
`1.8* 10^5\ N/C`

T = Tension

q = Charge

We have the equations

`Tcos\theta=mg`

`Tsin\theta=qE`

Dividing the equations

`tan\theta=(mg)/(qE)\\\Rightarrow q=(mgtan\theta)/(E)\\\Rightarrow q=(5* 10^(-3)* 9.81* tan23)/(1.8* 10^5)\\\Rightarrow q=1.15669* 10^(-7)\ C`

The magnitude of the charge is
`1.15669* 10^(-7)\ C`