Question

Two equally charged, 2.098 g spheres are placed with 3.338 cm between their centers. When released, each begins to accelerate at 269.429 m/s^2. What is the magnitude of the charge on each sphere?

Answer 1

`2.64* 10^(-7) C`

Step-by-step explanation:

There are two spheres name 1 and 2 and they posses the same charge, which is +q.

And they have equal mass which is 2.098 g.

The distance between these two spheres is,
`r=3.338 cm`.

And the acceleration of each sphere is,
`a=269.429 m/s^(2)`.

Now the coulumbian force experience by 1 sphere due to 2 sphere,

`F_(21) =(q^(2) )/(4\pi\epsilon_(0) r^(2)  )`.

And also the newton force will occur due to this force,

`F_(21)=ma`.

Now equate the above two values of force will get,

`(q^(2) )/(4\pi\epsilon_(0) r^(2)  ) =ma`

Further solve this,

`q^(2)=ma4\pi  \epsilon_(0) r^(2)`.

Substitute all the known variables in above equation,

`q^(2)=(2.098* 10^(-3) )(269.429)(4(3.14))(8.85* 10^(-12))(3.338* 10^(-2))`.

`q=2.64* 10^(-7) C`.