A regular hexagon of side 5 cm has charge `10 mu C` at each of its vetices . The potential at the centre of hexagon is

Options

A. o v
B. `18 xx 10^(5)V`
C. ` 1.08 xx 10^(7) v`
D. `1.08 xx 10^(5)v`

Answer

C

Explanation

Given
length of sides of hexagon r= 5 cm `=5 xx 10^(-2)` m
charge on each vertices of hexagon q=10 c
Now according to the question

Dut to one charge potential on hexagon centre
P is given potential `v= (1)/(4pi epsilon_( 0)) . (q)/(r )`
Due to all six charge total potential at the centre of hexagonal is `V_("total ") =6 xx (V_(1))`
Now from Eq. (i ) we get
`V_("total" ) = 6 xx 18 xx 10^(5) rArr V_("total" ) = 1.08 xx 10^(7) V`