Conisder a system of three charges `q//3`, `q//3` and `-2q//3` placed at point A, B and C, respectively, as shown in the figure. Take O to be centre of the circle of radius R and angle `CAB=60^@`

Options

A. The electric field at point O is `q/(8pi epsilon_(0)R^(2))` directed along the negative x-axis
B. The potential energy of the system is zero
C. The magnitude of the force between the charges at C and B is `(q^(2))/(54piepsilon_(0)R^(2))`
D. The potential at point O is `1/(12pi epsilon_(0)R^(2))`

Answer

C

Explanation

Electric field at point O is vector sum of electric field due to all three point charges individually.

`vecE_(@)=vecE_(a)+vecE_(B)+vecE_(c)" "` and `" "vecE_(A)=-vecE_(B)`
So `vecE_(@)=vecE_(c)`
`=(2q//3)/(4pi epsilon_(0)R^(2))` along `vec(OC)`
`=q/(6piepsilon_(0)^(2)R^(2))` along `vec(OC)`
`=-q/(6 pi epsilon_(0)R^(2))hati`
From geometry, we can find that `/_ABC=30^(@)`
and `/_ACB=90^(@)`
so `AB=2R,AC=R,BC=sqrt(3)R`
Total potential energy of the system is
`U=1/(4 pi epsilon_(0))[(q//3xxq//e)/(2R)-(q//3xx2q//3)/R-(q//3xx2q//3)/(sqrt(3)R)]!=0`
`F=(q//3xx2q//3)/(4pi epsilon_(0)xx(sqrt(3)R)^(2))`
`(q^(2))/(54piepsilon_(0)R^(2))`
Potential at point O is `V=(q//3+q//3-2q//3)/(4pi epsilon_(0)R)=0`