Question

Four particles each having charge q are placed at the vertices of a square a. The value of the electric potential at the midpoint of one of the side will be

Answer 1

PC = \(\sqrt{(a/2)^2+a^2}\)

PC = \(\sqrt{a^2/4+a^2}\)

PC = \(\frac{\sqrt5a}2\) 

Then electric potential

V = \(\frac{kq}{\sqrt5a/2}\) + \(\frac{kq}{\sqrt5a/2}\) + \(\frac{kq}{a/2}\) + \(\frac{kq}{a/2}\)

V = \(\frac{2kq}{\sqrt5a}\) + \(\frac{2kq}{\sqrt5a}\) + \(\frac{2kq}{a}\) + \(\frac{2kq}{a}\)

V = \(\frac{4kq}{\sqrt5a}\) + \(\frac{4kq}{a}\)

V = \(\frac1{4\pi\varepsilon_0}\frac{2q}a(\frac2{\sqrt5}+2)\) 

Comments

thank you so much ^^