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 :

Answer:

[tex]2.64\times 10^{-7} C[/tex]

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, [tex]r=3.338 cm[/tex].

And the acceleration of each sphere is, [tex]a=269.429 m/s^{2}[/tex].

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

[tex]F_{21} =\frac{q^{2} }{4\pi\epsilon_{0} r^{2}  }[/tex].

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

[tex]F_{21}=ma[/tex].

Now equate the above two values of force will get,

[tex]\frac{q^{2} }{4\pi\epsilon_{0} r^{2}  } =ma[/tex]

Further solve this,

[tex]q^{2}=ma4\pi  \epsilon_{0} r^{2}[/tex].

Substitute all the known variables in above equation,

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

[tex]q=2.64\times 10^{-7} C[/tex].

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