the excess(equal in no) of electrons that must be placed on each of two spheres spaced 3 cm apart, with force of repulsion between sphere to be 10*-19 is

Question

Arun · Answer

Dear pallavi

let the charge on spheres is q

so F= 1/4∏ε qq/.03²

given F= 10^-19

1/4∏ε qq/.03²=10^-19

9 *10⁹ * q² /.03²=10^-19

q=10^-16

1 electron has 1.6 * 10^-19 C

ne =q

n=q/e =10^-16/1.6 * 10^-19 C

=625

Regards

Arun

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the excess(equal in no) of electrons that must be placed on each of two spheres spaced 3 cm apart, with force of repulsion between sphere to be 10*-19 is

the excess(equal in no) of electrons that must be placed on each of two spheres spaced 3 cm apart, with force of repulsion between sphere to be 10*-19 is

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