Question

Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each other at 414 m/s2. What is the magnitude of the charge on each sphere, assuming only that the electric force is present?

Answer 1

Charge,
`q=1.35* 10^(-7)\ C`

Step-by-step explanation:

It is given that,

Mass of spheres, m = 1 g = 0.001 kg

Distance between spheres, r = 2 cm = 0.02 m

Acceleration when they released,
`a=414\ m/s^2`

We need to find the magnitude of the charge on each sphere. The electric force is given by :

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

Also, F = ma

`ma=(kq^2)/(r^2)`

`q^2=(mar^2)/(k)`

`q^2=(0.001* 414* (0.02)^2)/(9* 10^9)`

`q=1.35* 10^(-7)\ C`

So, the magnitude of charge on each spheres is
`1.35* 10^(-7)\ C`. Hence, this is the required solution.

Answer 2

The magnitude of the charge on each sphere is 0.135 μC

Step-by-step explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

We need to calculate the magnitude of charge

Using newton's second law

`F= ma`

`a=(F)/(m)`

Put the value of F

`a=(kq^2)/(mr^2)`

Put the value into the formula

`414=(9*10^(9)* q^2)/(1.0*10^(-3)*(2.0*10^(-2))^2)`

`q^2=(414*1.0*10^(-3)*(2.0*10^(-2))^2)/(9*10^(9))`

`q^2=1.84*10^(-14)`

`q=0.135*10^(-6)\ C`

`q=0.135\ \mu C`

Hence, The magnitude of the charge on each sphere is 0.135μC.