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

Options

A. The electric field at point O is `q/(8piepsilon_0R^2)` directed along the neagtive 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_0R^2)`
D. The potential at point O is `q/(12piepsilon_0R^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)/(4piepsilon_0R^2)` along `vecOC`
`=q/(6piepsilon_0^2 R^2)` along `vecOC`
`=-q/(6piepsilon_0R^2) hati`
From geometry , we can find that `angleABC=30^@` and `angleACB=90^@`
So, AB=2R , AC=R , BC=`sqrt3R`
Total potential energy of the system is,
`U=1/(4piepsilon_0)[(q//3xxq//3)/(2R) -(q//3xx2q//3)/R-(q//3xx2q//3)/(sqrt3R)]ne0`
`F=(q//3xx2q//3)/(4piepsilon_0xx(sqrt3R)^2)`
`=q^2/(54piepsilon_0R^2)`
Potential at point O is , `V=(q//3+q//3-2q//3)/(4piepsilon_0R)=0`