Question

Four charges are placed at the
corners of a square of side a. ABD are +q and C is -q. find net force on the charge at corner C​

Answer 1

Answer:

F = KQq/r²

hence , magnitude of force depends upon products of magnitude of charges

and distance between them .

but here distance between charge particle is fixed so,

only depends upon product of magnitude of charge .

we know smallest charge = 1.6 × 10-19C

of electrons exist in nature .

so, if we choose both charge particle " electrons " then force will be minimum.

now,

F = Ke²/r²

= 9× 10^9 × (1.6 × 10^-19)²/(1/100)²

= 9 × 2.56 × 10^-29+4 N

= 23.04 × 10^-25 N

= 2.304 × 10^-24 N

4.6