Question

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 potential energy of the system in first situation where the charge is divided into three equals parts is
A. `(1)/(4piepsilon_(o))(q^(2))/(r)`
B. `(1)/(36piepsilon_(o))(q^(2))/(r)`
C. `(1)/(12sqrt(3)piepsilon_(o))(q^(2))/(r)`
D. `(1)/(12piepsilon_(o))(q^(2))/(r)`

Answer 1

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)`