Find the work done in placing a charge Q/2 at . . .

Question : Find the work done in placing a charge Q/2 at point E in the given square configuration.




Doubt by Mansi 

Solution : 

AB=BC=CD=DA = a 
BE=QE=a/2
CE=DE = √(a)²+(a/2)²
[By Pythagoras Theorem]

= √(a)²+(a/2)²
= √a²+(a²/4)
= √[(4a²+a²)/4]
= √[5a²/4]
= a√5/2 
CE=DE = a√5/2 

W = W1+W2+W3+W4
W =kQ²/a + kQ²/a + kQ²/a√5 + kQ²/a√5
W = kQ²/a[1+1+(1/√5)+(1/√5)]
W = kQ²/a[2+2/√5]
W = 2kQ²/a[1+1/√5]
W = 2kQ²/a[(√5+1)/√5]
W = 2kQ²/a[(5+√5)/5]
W = 2kQ²(5+√5)/5a