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) ⇒= ⇒= ⇒=()2 ⇒2= ⇒2x=x+ ⇒ x = 5 cm