Question

Two small objects are hung, from a point, on strings 50 cm long. they each have mass .06 grams and are given equal charges if the strings each make an angle of 37 degrees with a vertical. What is the magnitude of the electric force between the objects? What is the magnitude of the charge on each object?

Answer 1

(a). The magnitude of the electric force between the objects is
`4.43*10^(-4)\ N`

(b). The magnitude of the charge on each object is 0.133μC.

Step-by-step explanation:

Given that,

length of string = 50 cm

Mass of each object= 0.06 g

Angle = 37°

Let the electric force magnitude be F.

So, the balancing forces perpendicular to the string,

We need to calculate the magnitude of the electric force between the objects

Using formula of force

`mg\sin\theta=F\cos\theta`

Put the value into the formula

`0.06*10^(-3)*9.8\sin37=F\cos37`

`F=(0.06*10^(-3)*9.8\sin37)/(\cos37)`

`F=0.0004430\ N`

`F=4.43*10^(-4)\ N`

We need to calculate the distance

Using formula of distance

`\sin37=((d)/(2))/(50)`

`d=\sin37*100`

`d=60.18\ cm`

We need to calculate the magnitude of the charge on each object

Using formula of electric force

`F=(kq^2)/(d^2)`

Put the value into the formula

`4.43*10^(-4)=(9*10^(9)* q^2)/((60.18*10^(-2))^2)`

`q^2=(4.43*10^(-4)*(60.18*10^(-2))^2)/(9*10^(9))`

`q=\sqrt{(4.43*10^(-4)*(60.18*10^(-2))^2)/(9*10^(9))}`

`q=0.133*10^(-6)\ C`

`q=0.133\ \mu C`

Hence, (a). The magnitude of the electric force between the objects is
`4.43*10^(-4)\ N`

(b). The magnitude of the charge on each object is 0.133μC.