At each of the four vertices of a square of side 10 cm, four positive charges each of 20 esu are placed. Find the potential at the point of intersection of the two diagonals.

Explanation

Amount of charge at each corner, q = 20 esu of charge
Length of each side of the square = 10 cm
So, length of diagonal = `sqrt(10^(2) + 10^(2)) = 10sqrt(2)cm`
`:.` The distance of a vertex from the point of intersection of the diagonals is,
`:. x = 1/2 xx "length of diagonal" = 1/2 xx 10sqrt(2) = 5sqrt(2) cm`
`:.` Potential at the point of intersection of the two diagonals
`=(20)/(5sqrt(2)) + (20)/(5sqrt(2)) + (20)/(5sqrt(2)) + (20)/(5sqrt(2))`
`(80)/(5sqrt(2)) = 8sqrt(2) = 11.31 stat V`.