Here, AC = BC = 2a D and E are the midpoints of BC and AC ∴ AE = EC = a and BD = DC = a In ΔADC,(AD)2=(AC)2−(DC)2=(2a)2−(a)2=4a2−a2=3a2 AD=a√3 Similarly, BE=a√3 Potential at point D due to the given charge configuration is VD=
1
4πε0
[
q
BD
+
q
DC
+
q
AD
] =
q
4πε0
[
1
a
+
1
a
+
1
√3a
]=
q
4πε0a
[2+
1
√3
]...........(i) Potential at point E due to the given charge configuration is VE=
1
4πε0
[
q
AE
+
q
EC
+
q
BE
] =
q
4πε0
[
1
a
+
1
a
+
1
a√3
]=
q
4πε0a
[2+
1
√3
].....(ii) From the (i) and (ii), it is clear that VD=VE The work done in taking a charge Q from D to E is W=Q(VE−VD)=0(∵VD=VE)