Solving Electric Problem: Q/q & Net Force on Particles

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In summary, Q has a negative charge and q has a positive charge. When the net electrostatic force on particles 1 and 4 is zero, Q = -2q/a.
  • #1
KyoPhan
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My semester just started and I'm already struggling.

The diagram is a square with a dot on each corner representing a charged particle. 1 at top left, 2 at top right, 3 at bottom left, and 4 at bottom right. The distance on one side of the square is a
The problem reads In fig. 21-22, four particles form a square. The charges are q1 = q4 = Q and q2 = q3 = q.

(a) What is Q/q if the net electrostatic force on particles 1 and 4 is zero?

(b) Is there any value of q that makes the net electrostatic force on each of the four particles zero? Explain


For (a)
I think I have to apply Coulomb's law and calculate all the forces from each individual particle. Assuming that 1/4 are both negative and 2/3 are positive, I calculated the force upward (by adding the force caused between 1/4 and 1/3) on particle 1. Then I set them equal to 0. I solved for Q in relation with q and got Q = -2q/a . So I calculated Q/q and got -2/a.

For (b)
I want to say that q=Q but one of opposite sign because I remember my professor saying that if one charge is in equilibrium, then the rest are at well. (correct me if I'm using the terms or concept incorrectly). Or is this only when they are semetric because I remember him talking about it when there was a square with an electron on each corner, with a proton in the middle. Do I have the right idea or am I completely lost? Do I have to apply Coulomb's equation in some way?

Thanks you for taking your time, I really appreciate it.
 
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  • #2
Well, since Q and q are both quantities of charge, in [Coulomb],
is it possible that "a" has no units? Is "a" a distance, in [meter] ?

The horizontal component of Force on 1 must be zero ... so that :
kQq/a^2 = {kqq/(1.414 a)^2}sin 45 => Q/a^2 = .707q/(2a^2) ...
 
  • #3
A has no units, its just an unknown distance

I'm sorry I'm kinda slow, but where did 1.414a come from?
 
  • #4
KyoPhan said:
A has no units, its just an unknown distance

I'm sorry I'm kinda slow, but where did 1.414a come from?
It's the square root of two. Since we're dealing with a square, the length of the diagonal is [tex]s\sqrt{2}[/tex], where [tex]s[/tex] is the side length.
 
  • #5
lightgrav said:
Well, since Q and q are both quantities of charge, in [Coulomb],
is it possible that "a" has no units? Is "a" a distance, in [meter] ?

The horizontal component of Force on 1 must be zero ... so that :
kQq/a^2 = {kqq/(1.414 a)^2}sin 45 => Q/a^2 = .707q/(2a^2) ...

O okay I understand it now, thanks a lot

sorry for late reply btw
 

1. What is the difference between Q and q in solving electric problems?

Q and q refer to different quantities in electric problems. Q typically represents a fixed charge, while q represents a variable charge. Q is usually associated with larger objects or systems, while q is used for smaller particles or charges.

2. How do you calculate the net force on particles in an electric field?

To calculate the net force on particles in an electric field, you can use the equation F = qE, where F is the net force, q is the charge of the particle, and E is the electric field strength. Simply multiply the charge of the particle by the electric field strength to find the net force.

3. Can you solve electric problems without knowing the net force on particles?

Yes, it is possible to solve electric problems without knowing the net force on particles. You can use other equations, such as the equation for electric potential energy (U = qV), to find the solution. However, knowing the net force can provide valuable information about the behavior of particles in an electric field.

4. How do you handle multiple charges in an electric problem?

To handle multiple charges in an electric problem, you can use the principle of superposition. This means that the total electric field at a point is the vector sum of the individual electric fields from each charge. You can also use the principle of superposition to find the net force on a particle due to multiple charges.

5. What is the importance of units in solving electric problems?

Units are crucial in solving electric problems because they provide a standardized way to measure and compare quantities such as charge, electric field, and force. It is important to use the correct units and to convert between units when necessary to ensure accurate solutions to electric problems.

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