A charge q is divided into three equal parts and placed symmetrically on a circle of radius r. The same charge is divided into four equal parts and placed symmetrically on the same circle. The electric field intensities at the centre of the circle in two situations are zero
Q. The ratio of electric potentials at the centre in the two situations is

Options

A. `(2)/(sqrt(3))`
B. `(1)/(1)`
C. `(4)/(3)`
D. `(16)/(9)`

Explanation

Potentials at the centre
`V_(1)=(1)/(4piepsilon_(0))(q)/(r), V_(2)=(1)/(4piepsilon_(0))(q)/(r)`
Potential energy in situation `I` is
`U_(1)=3xx(1)/(4piepsilon_(0))((9//3))^(2)/((sqrt(3)R))` `=(1)/(12sqrt(3)piepsilon_(0))(q^(2))/(R)`
When one charge is removed, the field intensity at the centre is due to the removed charge only.
`E_(1)=(1)/(4piepsilon_(0))(q//3)/(r^(2))`
`E_(2)=(1)/(4piepsilon_(0))(q//4)/(r^(2)) therefore(E_(1))/(E_(2))=(4)/(3)`