Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown in the figure. Given that `K=(1)/(4piepsilon_0)(q)/(L^2)`, which of the following statements(s) is (are) correct?

Options

A. The electric field at `O` is `6K` along `OD`
B. The potentail at `O` is zero
C. The potentail at all points on the line `PR` is same
D. The potential at all points on the line `ST` is same

Answer

A, B, C

Explanation

In Fig, `L` is the length of each side of a regular hexagon `ABCDEF`, at the vertices of which six point charges are held.

Imagine a unit positive charge at `O`.
Electric intensity at `O` due to `+2q` charge at `A`
`= (2q xx 1)/(4pi in_(0) L^(2)) = 2K`, along `OD`
Electric intensity at `O` due to `-2q` charge at `D = 2 K`, along `OD`.
Electric intensity at `O` due to `+q` charge at `B` and `-q` charge at `E = 2 K`, along `OE`
Electric intensity at `O` due to `-q` charge `C` abd `+q` charge at `F = 2K`, along `OC`
As `/_COE = 120^(@)`, therefore , resultant of `2K` along `OC` and `2K` along `OD`.
`:.` Net electric intensity at `0 = 4K + 2K = 6K` along `OD`
Potential at `O, V = - (dE)/(dr) = (d)/(dr) (6K) = Zero`
A closer look at figure shows that `PR` is equatorial line of these electric dipoles `(-q, +q)` at `E, F , (-2q +2q)` at `D, A` and `(-q, +q)` at `C, B`.
Hence potential at all points on `PR` is zero (i.e., same).
The choices (a), (b), (c ) are correct.