Class 12 NEET Physics Electrostatics Questions Part 1

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)