Two points charges + 9e and +e are kept 16 cm apart to each other. Where a third charge q is placed between them so that the system is in the equilibrium state

Question

in simple

Answer 1

In order for the system to be in equilibrium, the net force on the third charge q must be zero. This means that the electric force on q due to the charge +9e must be exactly opposite and equal in magnitude to the electric force on q due to the charge +e.

The electric force between two charges is given by Coulomb's law:
F = k * (|q1| * |q2|) / r^2
where F is the electric force, k is the Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.

Let's calculate the forces on q due to the charges +9e and +e.

Force due to +9e:
F1 = k * (|9e| * |q|) / r1^2

Force due to +e:
F2 = k * (|e| * |q|) / r2^2

Since the distances r1 and r2 are not given, we can't calculate the exact values of the forces. However, we can derive a relationship between r1 and r2 that results in a net force of zero on q.

Since the system is in equilibrium, the magnitudes of the forces must be equal:
F1 = F2

k * (|9e| * |q|) / r1^2 = k * (|e| * |q|) / r2^2

Simplifying the equation gives:
9e / r1^2 = e / r2^2

Cross multiplying gives:
e * r1^2 = 9e * r2^2

Simplifying further:
r1^2 = 9 * r2^2

Taking the square root of both sides gives:
r1 = 3 * r2

Therefore, in order for the system to be in equilibrium, the distance of the third charge q from the charge +9e must be three times the distance of q from the charge +e.