i need help with physics

Question

Two point charges q and 4q are at x=0 and x=L, respectively, and free to move. A third charge is placed so that the entire three-charge system is in static equilibrium.

What is the x-coordinate of the third charge?

Answer 1

Let

s be the x-position of the third charge (to be calculated),

Q be the amount of the third charge (unknown),

k be Coulomb's constant.

First consider the polarity of Q and its position s relative to the other two charges.

Without the third charge, the two given charges repulse each other, as they have the same polarity.

To stop the repulsion, the third charge must have the opposite polarity, and

must be placed between the two charges;

that is, 0 < s < L. (1)

You can eliminate all the other possibilities of placing the third charge by considering them.

Having decided on the position s relative to the other two,

we can now restrict ourselves to only the absolute values of the the resulting forces.

Let

F1 = k|q||Q|/s² be the magnitude of the electrostatic force between the charge q and Q.

F2 = k|4q||Q|/(L-s)² be the magnitude of the electrostatic force between the charge 4q and Q.

F3 = k|q||4q|/L² be the magnitude of the electrostatic force between the charge q and 4q.

Each of the three charges has a pair of forces acting on it.

We have arranged the position of Q so that the forces in each pair act in opposite direction.

Therefore we achieve equilibrium if the forces in each pair have the same magnitude.

That gives us three equations;

F1 = F2

F1 = F3

F2 = F3

It will turn out that the first equation is sufficient to determine s.

If your teacher also wanted you to calculate Q, you could do that by plugging the

(bellow) calculated value of s into the second or third equation.

F1 = F2

k|q||Q|/s² = k|4q||Q|/(L-s)²

1/s² = 4/(L-s)²

(L-s)² = 4s²

3s² + 2Ls - L² = 0

s = (-2L ± √((2L)² + 12L²))/6 = (-2L ± 4L)/6

That gives two solutions

s = L/3

s = -L

We can reject the second solution because of (1), so the answer is s = L/3.