Calculation of the net force on a test charge

[Quanta]

Homework Statement

Homework Equations

\overrightarrow{F} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2}\hat{r}

The Attempt at a Solution

I found to hard summarizing all forces directly adding each to other.

The solution of this problem is only matter of drawing the 13-sided polygon and forces acting on a test charge and summarizing them or there is something else ?

I think there is a trick to solve this...

 

[haruspex]

think there is a trick to solve this...

Quite so. Unfortunately it is hard to give a hint that does not equate to being the solution.
Maybe think of it from the perspective of different q charges. Why would Q move towards one rather than towards another?

 

[kuruman]

Suppose you were to add the 13 vectors graphically by putting them tail to tip as you would with any graphical vector addition ...

 

[Quanta]

Maybe think of it from the perspective of different q charges. Why would Q move towards one rather than towards another?

It seems, I have come to the right answer. If I take one particular q charge, all other charges are placed in the same way as if another q charge was chosen, because the 13-sided polygon has a symmetry. So net force on a test Q charge will be zero. am I right ?

 

[haruspex]

It seems, I have come to the right answer. If I take one particular q charge, all other charges are placed in the same way as if another q charge was chosen, because the 13-sided polygon has a symmetry. So net force on a test Q charge will be zero. am I right ?

Right. kuruman's approach is good too.

 

[Quanta]

But there is another problem that relates to this. Here is :

In this case charges are not symmetrically placed as in previous example. But we have one symmetry if we draw a line from the Q charge to the missing q charge.

I have an answer :

Why towards the missing q ? Why exactly \frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2} ?

 

[haruspex]

But there is another problem that relates to this. Here is :

View attachment 213485
In this case charges are not symmetrically placed as in previous example. But we have one symmetry if we draw a line from the Q charge to the missing q charge.

I have an answer :

View attachment 213486

Why towards the missing q not opposite to it ? Why exactly \frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2} ?

Can you think of some other simple change that would have been equivalent to removing one charge q?

 

[Quanta]

Can you think of some other simple change that would have been equivalent to removing one charge q?

I tried with 5 sided regular polygon, removing one charge in the corner and adding graphically all force vectors I got net force vector directed toward missing charge. The same must be applied with 13 sided regular polygon without adding all vectors, from the figure we see that there is a symmetry represented by blue axis, forces from left and right sides cancel each other, so Q should move toward the missing charge, because of gap between charges (and because of the more dense placement on the south side ?). The magnitude of force on Q is \frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2} where r = r_1 + r_2 + ... + r_{12}. Is everything ok ? Any corrections ?

 

[Quanta]

I already know answer from the solution manual :

I want to get this answer with the help of logical reasoning without using any geometric properties of adding all vectors.

 

[haruspex]

I tried with 5 sided regular polygon, removing one charge

Ok, but I said instead of removing a charge. What, equivalently, could you add?

 

[Quanta]

I have an explanation. When 13 charges were present the system was balanced, the applied force on Q from any charge is \frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2} this is a force to maintain equilibrium, after removing charge system becomes unbalanced, so the net force points toward the missing charge and is \frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2}.

What, equivalently, could you add?

add charge(s)... but where ? I'm interesting where you're pointing me.

 

[haruspex]

add charge(s)... but where ?

If you have a charge q at point P, what charge could you add which is electrically equivalent to removing the charge q?

 

[Quanta]

If you have a charge q at point P, what charge could you add which is electrically equivalent to removing the charge q?

Opposite to it.

 

[haruspex]

Opposite to it.

Yes, add a charge of -q at P. We know that the q charges balance each other at the centre, so what is the effect there of adding the -q charge?

 

[Quanta]

We know that the q charges balance each other at the centre, so what is the effect there of adding the -q charge?

System will become unbalanced and $Q$ will move toward the $-q$.

 

[haruspex]

System will become unbalanced and $Q$ will move toward the $-q$.

Assuming Q and q have the same sign, yes. But what is the magnitude of the force?

 

[Quanta]

But what is the magnitude of the force?

$Q$ and $-q$ have different signs, there will be attraction force between them with magnitude of $\frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2}$.

 

[haruspex]

$Q$ and $-q$ have different signs, there will be attraction force between them with magnitude of $\frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2}$.

Yes, except that you cannot say what the signs of Q and -q are. It could be that Q is positive and q is negative. Then Q and -q will have the same sign. However, you can write that there will be a force of attraction of $\frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2}$. If Q and q have the same sign then Q and -q will have opposite sign, so it is indeed attraction; if Q and q have opposite sign then the force will be a repulsion, but that's ok because now $\frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2}$ will be negative, and saying an attraction of $\frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2}$ is the same as saying a repulsion of |$\frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2}$|.

 

[Quanta]

Yes, except that you cannot say what the signs of Q and -q are...

I have no problems with signs, but thank you anyway. I appreciate your help.

 

[haruspex]

I have no problems with signs, but thank you anyway. I appreciate your help.

Well, you wrote

Q and −q have different signs

which suggests to me you do have a misunderstanding. If Q is +e and q is -e then Q and -q have the same sign: they are both +e.