Question

net force on central q charge?​

Answer 1

Answer:

$k\frac{2q^2}{r^2}$ in the vertical direction.

Explanation:

Net force on central charge q will be due to only the charge +2q. Rest all the charges are placed symmetrically and hence will cancel each other out.

The net force experienced by the charge q in the vertical direction (attraction)

$F=k\frac{2q\times q}{r^2}$

$\implies F=k\frac{2q^2}{r^2}$

Where k is Coulomb's constant.

Hope this helps.