Question

Some charges are kept on a circular dial so that 1 unit of charge is at the
1’O clock position, 2 units at 2’O clock position, and so on ending with 12
units of charge at the 12’Oclock, position. Find the magnitude and the
direction of the electric field at the center of the dial.

Answer 1

Answer:

if the charges are positive then the electric field is towards right side and vice versa

the magnitude of resulting electric field is:

2+√3(kq/r)

here r is radius of dial

.....

you figure it out by calculating electric field due to each charge and adding them by rule of vector addition