Question

5 equal charges i.E.+q are placed at 5 corner of regular hexagon of side



a.Find the magnitude of electric field at centre ?

Answer 1

Answer:

Electric field at Centre,

E=(1/4pi epsilonnot) q/r^2

=9*10^9 q/r^2

Explanation:

Two pairs will be canceled out due to opposite of each other. Only electric field will remain due to single charge.

Answer 2

Answer:

E = kq/a^2

Explanation:

go through the attachment.