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Two charges +4e and +e are at a distance...

Two charges `+4e` and `+e` are at a distance `x` apart. At what distance,a charge `q` must be placed from charge `+e` so that is in equilibrium

A

`x//2`

B

`2x//3`

C

`x//3`

D

`x//6`

Text Solution

Verified by Experts

The correct Answer is:
C
`F=1/(4 pi epsilon_(0)) K . (q_(1)q_(2))/(r^(2))`
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