Five point charge (`+q` each) are placed at the five vertices of a regular hexagon of side `2a`.what is the magnitude of the net electric field at the centre of the hexazon?

Options

A. 1. `(q)/(4 pi epsilon_(0)a^(2))`
B. 2. `(q)/(8 pi epsilon_(0)a^(2))`
C. 3. `(q)/(16 pi epsilon_(0)a^(2))`
D. 4. zero

Answer

A

Explanation

`E=(1)/(4 pi epsilon_(0)) (q)/(a^2)`
Suppose the charge is present at the sixth vertex also, then electric field at center would be zero. Now, if charge is not present at this vertex. The electric field at center would be because of other five charges, which should be equal and opposite to the fiedl produced due to single charge at the sixth vertex.