Question

A charge +Q is located at the origin and a second charge, +4Q, is at distance d on the x-axis. If a third charge, q, is placed somewhere, so that all three charges will be in equilibrium.1. What should be its sign, so that all three charges will be in equilibrium?A. Negative.B. Positive.2. What should be its magnitude, so that all three charges will be in equilibrium?

Answer 1

In the scenario with charges +Q, +4Q, and q, the charge q must be negative to achieve equilibrium. For the electric dipole case, the net force on a negative charge equidistant from the dipole will be towards the negative charge of the dipole.

Step-by-step explanation:

To determine the equilibrium conditions for a layout of electric charges, one must apply Coulomb's law, which dictates that the magnitude and direction of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In the first scenario, the third charge q must be negative to be in equilibrium with the two positive charges (+Q and +4Q), because it needs to be attracted to both. The exact placement and magnitude would depend on its distance from the charges.

Regarding the scenario with the electric dipole and third charge, the direction of the net force on the third charge, which is -q and equidistant from the dipole charges of +2q and -2q, would be towards the negative charge of the dipole. This is because the attractive force between unlike charges is stronger than the repulsive force between like charges when the distances are equal, due to the dipole's differing charge magnitudes.

Answer 2

Third charge should be placed at a distance d/3

Sign is negative

Magnitude is |q| = 4Q/3

Step-by-step explanation:

To find a place where a third charge, q, is placed so that all three charges will be in equilibrium would be at;

E1 = E2

Now, let the distance between the charges +Q and +4Q be d. Also, let the distance from charge +Q to the third charge q be x.

Thus;

E1 = kQ/x²

E2 = k(4Q)/(d - x)²

Since equilibrium, then;

E1 = E2 gives;

kQ/x² = k(4Q)/(d - x)²

k and Q will cancel out to give;

1/x² = 4/(d - x)²

Cross multiply to get;

(d - x)² = 4x²

Taking square root of both sides gives;

√(d - x)² = √4x²

Gives;

d - x = 2x

d = x + 2x

d = 3x

x = d/3

This is the point at which E1 = E2.

Since at equilibrium, it means that Fq + F2 = 0

Now, electric force formula is;

F = kq1*q2/r²

So; Fq = kqQ/x²

F2 = kQ(4Q)/d²

Equating Fq + F2 = 0 gives;

kqQ/x² + kQ(4Q)/d² = 0

kqQ/x² = -kQ(4Q)/d²

Like terms cancel out to get;

q/x² = -4Q/d²

q = - 4Qx²/d²

earlier we saw that x = d/3.

So, q = -4Q × (d/3)²/d²

q = -4Q × ⅓

q = - 4Q/3

Thus,it's sign is negative.

It's magnitude will be the positive value.

Thus, magnitude is;

|q| = 4Q/3