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

Answer:

$2.64\times 10^{-7} C$

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,

$F_{21} =\frac{q^{2} }{4\pi\epsilon_{0} r^{2} }$.

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

$F_{21}=ma$.

Now equate the above two values of force will get,

$\frac{q^{2} }{4\pi\epsilon_{0} r^{2} } =ma$

Further solve this,

$q^{2}=ma4\pi \epsilon_{0} r^{2}$.

Substitute all the known variables in above equation,

$q^{2}=(2.098\times 10^{-3} )(269.429)(4(3.14))(8.85\times 10^{-12})(3.338\times 10^{-2})$.

$q=2.64\times 10^{-7} C$.