Six point charge each of the magnitude Q are placed at the vertices of a regular hexagon of side a as shown in the figure Electric field intensity on the line passing through the centre o and perpendicular to the plane of the figure at a large distance `x (gt gt a)` from o is
`(" Let " .(1)/(4pi in_(0) ) = k)`

Options

A. `k xx (4Q a)/(x^(3))`
B. `k xx (2Q a)/(x^(3))`
C. `k xx (8Qa)/(x^(3)) `
D. ` 0`

Answer

A

Explanation

According to the question

Let OK = due to AB force at the point K is only along AB
i.e., `F_(k ) = (2kQ a)/((a^(2) + x^(2))^((3)/(2)))`
there are three such `F_(k)` so the net resultant of these three
`F_(net ) = (4kQ a)/( (a^(2) + x^(2))^((3)/(2))) = (4kQa)/( x^(3)(1+ (a^(2))/(x^(2)))^((3)/(2)))`
`F_(net) = (4kQa)/( x^(3)) :. (a^(2))/( x^(2))=0) (x gt gt a)`