Four charges are arranged at the corners of a square ABCD of side d as show in figure.
(a) Find the work required to put together this arrangement.
(b) A charge `q_(0)` is brought to the centre E of the square the four charges being held fixed at its corners. How much extra work is needed to do this ?

Explanation

Suppose sides of square ABCD are AB = BC =CD= DA = d. Hence AC= BD=d `sqrt(2)`
(i) Work needed to bring charge +q to A when no charge is present on square is `W_(1)=0`
(ii) Work needed to bring -q to B when +q is at A
`W_(2) = (kq(-q))/(AB) =-(kq^(2))/(d) `
(iii) Now work needed to bring charge +q to C when +q is at A and -q is at B
`W_(3)=(k(+q)(+q))/(AC)+((k-q)(+q))/(BC)`
`:. W_(3) =(kq^(2))/(dsqrt(2))-(kq^(2))/(d)`
(iv) Now work done to bring -q to D when +q is at A ,-q at B and +q at C ,
`W_(4)=(K(+q)(-q))/(AD)+(k(-q)(-q))/(BD)+(k(+q)(-q))/(CD)`
`W_(4) =-(kq^(2))/(d)+(kq^(2))/(dsqrt(2))-(kq^(2))/(d)`
Hence, work done to bring given charge on square,
`W=W_(1)+W_(2)+W_(3)+W_(4)`
`:. W=0-(kq^(2))/(d) +(kq^(2))/(dsqrt(2))-(kq^(2))/(d)+(kq^(2))/(dsqrt(2))-(kq^(2))/(d)`
`=(-4kq^(2))/(d)+(2kq^(2))/(dsqrt(2))`
`=(-kq^(2))/(d)[4-sqrt(2)]`
`W=(-q^(2))/(4piin_(0)d)[4-sqrt(2)]`
Hence, the work done depends only on the arrangement of the charges.
(b) E is the centre of square ABCD. Suppose, r is the distance of every vertex and hence
`r=(d)/(sqrt(2))` . If V is the potential at E .
`V=(k(+q))/(d//sqrt(2))+(k(-q))/(d//sqrt(2))+(k(+q))/(d//sqrt(2))+(k(-q))/(d//sqrt(2))`
`:. V=0`
`:.` The work done to bring charge `q_(0)` at the point E
`W= q_(0)xxV`
`= q_(0) xx0`
`:. W=0`
Hence, no work is required to bring any charge to the centre of square. But, not on how they are brought to square vertices.