Static equilibrium for three point charges

In summary: The solution involves finding the correct location and magnitude of a third charge to create a static equilibrium in the system. The conversation also mentions the use of electrostatic force equations and assumptions about the charges being point charges with no external forces acting on them. It is determined that the third charge must be placed between the other two charges and have a negative magnitude in order for the system to be in a static equilibrium. The final solution for the distance and magnitude of the third charge is provided.
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DeIdeal
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The result I got seemed sort of messy and I'm not sure I've gotten the basic idea correctly, so I thought I'd make it sure. I think this was the easiest problem we had to solve for charge distributions and point charges and whatnot, so I know for sure I'll have to practice more if I made any big mistakes. I'm especially uncertain of my reasoning.

Homework Statement



A charge of 2q is placed at the origin and a charge of 3q is apart from it by a distance L. Where should a third charge be placed, and how large would it be, so the system would be in a static equilibrium.

Homework Equations



Electrostatic force F=(1/4πε_0)*((q_1*q_2)/r^2)*û_r

The Attempt at a Solution



The force on each charge has to be zero in order for the system to be in a static equilibrium. It is assumed that the charges are point charges and there are no external forces acting on them. Let the magnitude of the unknown charge be Q and the distance between 2q and Q be r.

The forces on the particle with a charge of 2q:

F = F_3q+F_Q = 0
k*(2q*3q)/L^2)*û_L + k*(2q*Q)/r^2)*û_r = 0
2qk* [ (3q/L^2)*û_L + (Q/r^2)*û_r ] = 0
(3q/L^2) * û_L = (Q/r^2) * û_r

Which is possible only if û_L || û_r, as 3q/L^2 and Q/r^2 are both scalars (I didn't make a mistake with this one, right?). So the unit vectors are parallel and three charges have to be on the same line.

Static equilibrium is only possible when the charge Q is negative, and placed inbetween the two other charges (as there must be forces to two directions on each particle).

Solving the magnitudes of Q and r, as the direction of r is no longer needed. Q is |Q| for now on, and I've noted the directions of the forces, choosing the direction from 2q to 3q to be positive. Two equations are enough for the two variables Q and r.

The forces on the particle with a charge of 2q:

F = F_Q - F_3q = 0
<=> k*(2q*Q)/r^2 - k*(2q*3q)/L^2 = 0
<=> 2q*k*( Q/r^2 - 3q/L^2 ) = 0
<=> Q/r^2 = 3q/L^2
<=> Q = 3q *(r/L)^2

The forces on the particle with a charge of 3q:

F = F_2q - F_Q = 0
<=> k*(3q*2q)/L^2 - k*(Q*3q)/(L-r)^2 = 0
<=> 3q*k*( 2q/L^2 - Q/(L-r)^2 ) = 0
<=> 2q/L^2 = Q/(L-r)^2 // The Q has been solved for r
<=> 2q/L^2 = (3q *(r/L)^2) /(L-r)^2
<=> 2q = 3q * (r^2*L^2)/(L^2*(L-r^2))
<=> 2/3 = r^2/(L-r)^2
<=> 2(L-r)^2 = 3r^2
<=> 2L^2 - 4Lr + 2r^2 = 3r^2
<=> r^2 + 4Lr - 2L^2 = 0

Solving for r eventually gives

r = (√6-2)L (~= 0.449L, makes sense to me as Q should be closer to 2q than to 3q)

Solving Q with r solved

<=> Q = 3q * ((sqrt6-2)L/L)^2
<=> Q = 3q * (6 - 4√6 - 4)
<=> Q = 6q * (5 - 2√6) (~= 0.606q, no idea whether this makes any sense or not. And the actual Q is obviously ~= -0.606q, as this was |Q|.)

Answer: r = (√6-2)L, Q = -6q*(5 - 2√6)

I'm not a native English speaker, so I apologize for any mistakes. (I also apologize for the lack of Latex, I've never gotten around learning it.)
 
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I assume the charges 2q and 3q are fixed. Consider the electric field left of, right of, and in between the fixed charges. At those locations place a positive charge, which way does it go, or where can the electric field sum to zero? See,
 

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1. How do I determine the net force on a charge in static equilibrium?

In static equilibrium, the net force acting on a charge is equal to zero. This means that the forces in all directions must be balanced. To determine the net force, you can use the principle of vector addition. Add all the forces acting on the charge, taking into account their direction and magnitude. If the result is zero, then the charge is in static equilibrium.

2. Can the forces acting on a charge be in equilibrium if the charges are not at the same distance from each other?

Yes, the forces can still be in equilibrium even if the charges are not at the same distance from each other. This is because the magnitude of the force between two charges is inversely proportional to the square of the distance between them. Therefore, as long as the distances are in the correct proportion, the forces will still balance out.

3. How do I determine the direction of the net force on a charge in static equilibrium?

The direction of the net force on a charge in static equilibrium is determined by the direction of the individual forces acting on the charge. If the forces are acting in opposite directions, then the net force will be in the direction of the larger force. If the forces are acting in the same direction, then the net force will be in that same direction.

4. Can the charges be in static equilibrium if one of the charges is negative?

Yes, the charges can still be in static equilibrium even if one of the charges is negative. The direction of the force between two charges depends on the sign of the charges, but the magnitude of the force remains the same. As long as the forces are balanced, the charges will be in static equilibrium.

5. How does the distance between the charges affect the equilibrium of the system?

The distance between the charges affects the equilibrium of the system because it determines the magnitude of the force between the charges. As the distance increases, the force decreases, and vice versa. Therefore, the equilibrium of the system can be affected if the distance between the charges changes, as the forces may no longer be balanced.

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