Question

Six identical charges Q are placed at the vertices of a
regular hexagon of side a. The electric field on the line
passing through centre of hexagon and perpendicular
to the plane of figure as a function of distance x from
centre is maximum at x = Xo. Value of xo is​

Answer 1

Thus the value of the electric field is zero.

Explanation:

Electric field "E"  = 1 /  4πϵo  x Q / r^3 x r

The value of the electric field at point "P" which is located at distance "x" from "O" is given by:

Solution:

Let K = 1 / 4πϵo

d = √ a^2 + x^2

Then EA = KQ / d^3 [ Kx - OA ]

Now EB = - KQ / d^3 [ Kx - OB ]

EC = KQ / d^3 [ Kx - OC ]

ED = KQ / d^3 [ Kx - OD ]

EE = KQ / d^3 [ Kx - OE ]

EF = KQ / d^3 [ Kx - OF ]

Now E = KQ / d^3 [ 1 -1 + 1 - 1 + 1 - 1 ] + [ OA + OB + OC + OD + OE + OF]

E = 0

Thus the value of electric field is zero.