- #1
TheShrike
- 44
- 1
Hello. This isn't a homework, but rather my own self-study of a textbook (Franklin's Classical Electromagnetism), so if this is an inappropriate place for these types of questions let me know.
Four equal charges q are placed at the vertices of a square of side L. What is the magnitude of force on one of the charges?
Further, if the charges are released from rest in this configuration use this force to find the velocity a long time after the charges are released.
Coulomb's law extended to many charges:
[itex]\mathbf{F}=q\sum_i\frac{q_i(\mathbf{r}-\mathbf{r}_i)}{|\mathbf{r}-\mathbf{r}_i|^3}[/itex]
In order to find the magnitude of the force I simply carried out the sum and used Pythagoras' theorem on the result. It is possible from inspection of the square to determine the vector quantities describing the distances from the single charge to the other three. For definiteness I am considering the force on the upper right-hand charge.
So,
[itex]\mathbf{F}=q\Big\{\frac{qL\mathbf{i}}{L^3}+\frac{qL \mathbf{j}}{L^3}+\frac{q(L\mathbf{i}+L\mathbf{j})}{(\sqrt{2}L)^3}\Big\}\\
=\Big(\frac{q}{L}\Big)^2\Big(1+\frac{1}{(\sqrt{2})^3}\Big)( \mathbf{i}+\mathbf{j})[/itex]
Then the magnitude of this force is
[itex]|\mathbf{F}|=\sqrt{2\Big(\frac{q}{L}\Big)^4\Big(1+ \frac{1}{2^{\frac{3}{2}}}\Big)^2}\\
=\Big(\frac{q}{L}\Big)^2\Big(\frac 12 +\sqrt{2}\Big)[/itex]
So assuming that is correct I continue, and here is where my trouble begins. Intuitively, the charges accelerate away from one another along the diagonals of the square. The force diminishes over time and so eventually the charges approach some final velocity.
I begin with Newton's #2:
[itex]\mathbf{F}=\Big(\frac{q}{L}\Big)^2\Big(1+\frac{1}{(\sqrt{2})^3}\Big)( \mathbf{i}+\mathbf{j})\\
=m\frac{d^2\mathbf{r}}{dt^2}[/itex]
The motion is identical in the x and y directions, so it suffices to consider motion in, say, the x direction.
[itex]\frac{d^2x}{dt^2}=\Big(\frac{q^2}{mL^2}\Big)\Big(1+\frac{1}{(\sqrt{2})^3}\Big)[/itex]
This is where I start to feel a little shaky in my logic. I want to say that L is now the initial separation of the charges along a certain direction. That is, taking the upper right-hand charge, L describes the initial separation between it and the charge to its left. Then I can integrate once to find the velocity. However this leads to a linear relationship between the velocity and time, which means that the velocity increases without bound.
Any thoughts?
Thanks in advance.
P.S. I can't figure out why some of the LaTeX code is not formatting properly. I hope it's clear regardless.
Homework Statement
Four equal charges q are placed at the vertices of a square of side L. What is the magnitude of force on one of the charges?
Further, if the charges are released from rest in this configuration use this force to find the velocity a long time after the charges are released.
Homework Equations
Coulomb's law extended to many charges:
[itex]\mathbf{F}=q\sum_i\frac{q_i(\mathbf{r}-\mathbf{r}_i)}{|\mathbf{r}-\mathbf{r}_i|^3}[/itex]
The Attempt at a Solution
In order to find the magnitude of the force I simply carried out the sum and used Pythagoras' theorem on the result. It is possible from inspection of the square to determine the vector quantities describing the distances from the single charge to the other three. For definiteness I am considering the force on the upper right-hand charge.
So,
[itex]\mathbf{F}=q\Big\{\frac{qL\mathbf{i}}{L^3}+\frac{qL \mathbf{j}}{L^3}+\frac{q(L\mathbf{i}+L\mathbf{j})}{(\sqrt{2}L)^3}\Big\}\\
=\Big(\frac{q}{L}\Big)^2\Big(1+\frac{1}{(\sqrt{2})^3}\Big)( \mathbf{i}+\mathbf{j})[/itex]
Then the magnitude of this force is
[itex]|\mathbf{F}|=\sqrt{2\Big(\frac{q}{L}\Big)^4\Big(1+ \frac{1}{2^{\frac{3}{2}}}\Big)^2}\\
=\Big(\frac{q}{L}\Big)^2\Big(\frac 12 +\sqrt{2}\Big)[/itex]
So assuming that is correct I continue, and here is where my trouble begins. Intuitively, the charges accelerate away from one another along the diagonals of the square. The force diminishes over time and so eventually the charges approach some final velocity.
I begin with Newton's #2:
[itex]\mathbf{F}=\Big(\frac{q}{L}\Big)^2\Big(1+\frac{1}{(\sqrt{2})^3}\Big)( \mathbf{i}+\mathbf{j})\\
=m\frac{d^2\mathbf{r}}{dt^2}[/itex]
The motion is identical in the x and y directions, so it suffices to consider motion in, say, the x direction.
[itex]\frac{d^2x}{dt^2}=\Big(\frac{q^2}{mL^2}\Big)\Big(1+\frac{1}{(\sqrt{2})^3}\Big)[/itex]
This is where I start to feel a little shaky in my logic. I want to say that L is now the initial separation of the charges along a certain direction. That is, taking the upper right-hand charge, L describes the initial separation between it and the charge to its left. Then I can integrate once to find the velocity. However this leads to a linear relationship between the velocity and time, which means that the velocity increases without bound.
Any thoughts?
Thanks in advance.
P.S. I can't figure out why some of the LaTeX code is not formatting properly. I hope it's clear regardless.
Last edited: