Question

Two equal in magnitude, positive charges Q are fixed at distance 2d from each other. The third charge, of magnitude q and unknown sign is positioned very carefully into the midpoint on the line connecting the first two charges. Pick the statement which is correct:

a. The third charge will accelerate towards one of the charges.
b. The third charge will be in equilibrium and this equilibrium is stable.
c. The third charge will be in equilibrium and this equilibrium is unstable.
d. The third charge will be in equilibrium. This equilibrium is stable if charge is positive and unstable if charge is negative.
e. All these statements are wrong.

Answer 1

the correct one is d

Step-by-step explanation:

To answer which statement is correct, let's look for the force on the third charge

F = F₁₃ + F₂₃

force is electric force

F =
`k (q_1 q_2)/(r^2)`

In the exercise we are told that the distance between the two charges is 2d, therefore the distance of each charge to the third is x = d and the two falls have a car Q

F₁₃ = k q Q / d²

F₂₃ = k q Q / d²

suppose that charge 3 is positive for which the forces are repulsive

F = k q Q / d₂ - k q Q / d₂

F = 0

suppose that the charge 3 is negative for which the forces are of attraction

F = -k q Q / d₂ + k q Q / d₂

F = 0

We can see that regardless of the sign of charge 3 the forces on it are zero, so it has no acceleration, it is in equilibrium.

Let's analyze the type of equilibrium, for this we move the charge 3 towards one of the sides

charge 3 positive

the force towards the approaching side increases and the other decreases, therefore the load has a net force towards the center, in that case the equilibrium is stable

negative charge 3

The force towards the approaching side is stronger and more attractive and the other decreases, therefore it has an attractive net force that carries it towards the charge, it is closer, in this case the balance is unstable

To review the different answers, the correct one is d