Class 12 NEET Physics Electrostatics Questions Part 1

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.