Four identical point charges q are placed at four corners of a square of side a. Find the potential energy of the charge system.
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Explanation

Method 1 (using direct formula)
`U = U_(12) + U_(13) + U_(14) + U_(23) + U_(24)+ U_(34)`
=`(K q^2)/a + (K q^2)/(a sqrt(2)) + (K q^2)/a + (K q^2)/a +(K q^2)/(asqrt(2))+ (K q^2)/a`
=`[(4 K q^2)/a +(2 K q^2)/(a sqrt(2))] = (2 K q^2)/a [2 + (1)/(sqrt(2))] = (a^2)/(2 pi epsilon _0 a )[2 + (1)/(sqrt (2))]`
Method 2 [using `U = (U_1 + U_2 + ...)]`
`U_1` is the total potential energy of charge at corner 1 due to all other charges , `U_2` is the total potential energy of charge at corner 2 due to all other chargers , `U_3` is the total potential energy of charge at corner 3 due to all other charges , `U_4` is the total potential energy of charge at corner 4 due to all other charges.
Since due to symmetry,
`U_1 = U_2 = U_3 =m U_4`
`U_("net") = (U_1 + U_2 + U_3 + U_4)/(2)`
`(1)/(2) . 4. [(K q^2)/a +(K q^2)/a +(K q^2)/(sqrt(2 a))] = (2 K q^2)/a [2+ (1)/(sqrt(2))]`.