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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
Dear pallavilet the charge on spheres is qso F= 1/4∏ε qq/.032 given F= 10-19 1/4∏ε qq/.032 =10-19 9 *109 * q2 /.032 =10-19 q=10-161 electron has 1.6 * 10-19 C ne =q n=q/e =10-16/1.6 * 10-19 C =625 RegardsArun
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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