From figure, AC=L,BC=L,BD=BC=L AD=AB+BD=2L+L=3L Potential at C is given by VC=
1
4πε0
[
q
AC
+
(−q)
BC
]=
1
4πε0
[
q
L
−
q
L
]=0 Potential at D is given by VD=
1
4πε0
[
q
AD
+
(−q)
BD
]=
1
4πε0
[
q
3L
−
q
L
]=−
q
6πε0
Work done in moving charge +Q along the semicircle CRD is given by W=[VD−VC](+Q)=[
−q
6πε0
−0](Q)=
−qQ
6πε0L
Comments : Potential at C is zero because the charges are equal and opposite and the distances are the same. Potential at D due to −q is greater than that at A(+q), because D is closer to B. Therefore it is negative.