AMU Engineering 2018 solved paper Solved Paper

Four charges having magnitude (−Q) are placed on the four corner of square having side ' a′ each.
q be the charge placed on centre ' O′ Electrostatic force between charges at A and B
FAB=

1

4πε0

.

Q2

a2

Similarly,
FBC=

1

4πε0

.

Q2

a2

and FDB=

1

4πε0

.

Q2

(√2a)2

[∵DB=√a2+a2=√2a]
FDB=

1

4πε0

.

Q2

2a2

Total force on the charge (−Q) at B due to charge at A,C and D.
F2=F1+FDB
∴ (where, F1 is the resultant force of FAB and FBC )
=√FAB2+FBC2+

1

4πε0

.

Q2

2a2

F2=√(

1

4πε0

.

Q2

a2

)2+(

1

4πε0

.

Q2

a2

)2 +

1

4πε0

.

Q2

2a2

F2=

1

4πε0

.

√2Q2

a2

+

1

4πε0

.

Q2

2a2

=

1

4πε0

.

Q2

a2

[√2+

1

2

]
=

1

4πε0

.

Q2

a2

[

2√2+1

2

] .........(i)
Electrostatic force between charge q at O and (−Q) at B
FOB=

1

4πε0

.

Qq

( √2a 2 )2

=

1

4πε0

.

Qq

a2 2

FOB=

1

4πε0

.

2Qq

a2

.........(ii)
Since, system of charge are in equilibrium, hence
FOB =F2

1

4πε0

.

2Qq

a2

=

1

4πε0

.

Q2

a2

(

2√2+1

2

)
2q=Q(

2√2+1

2

)
q=

Q

4

(2√2+1)