JEE Main 31-Aug-2021 Shift 1 Solved Paper

Charge of particle A,qA=20μC
Charge of particle B,qB=−5μC
Separation between A and B,R=5‌cm=5×10−2
The given situation is shown below. Here, qA=20μC,qB=−5μC,r=5‌cm
Let, third charge particle of charge qC is placed at point C to the right of charge B, so that net electric force would be equal to zero. This is because of following reason.
As the net electric force on particle C should be equal to zero, so the forces due to charged particles A and B must be in opposite direction. Hence, the particle should be placed on the line AB. As particles A and B having charges of opposite signs, so charged particle C cannot be between A and B.
Also, A has larger magnitude of charge than B. Hence, C should be placed closer to B than A.
Now, let position of C is at distance x from right of B.
∴ Force on C due to B(FCB)= Force on C due to A(FCA)
⇒

kqBqC

x2

=

kqAqC

(x+ 5 100 )2

⇒

(x+ 5 100 )2

x2

=

qA

qB

⇒

20

5

=(

x+ 5 100

x

)2
⇒2=

x+ 5 100

x

⇒2x=x+

5

100

⇒ x = 5 cm