Net force acting on a charged particle ##+Q##

[Mutatis]

Homework Statement

Twelve equal particles of charge $+q$ are equally spaced over a circumference (like the hours in a watch) of radius R. At the center of the circumference is a particle with charge $+Q$.

a) Describe the net force acting over $+Q$.

b) If the charge located at "3'o'clock" has been taken off, what's the net force acting over $+Q$?

c) If both charges located at "3'o'clok" and "9'o'clock" has been taken off, describe the force acting over +Q this time.

Homework Equations

$\vec F= \frac {1} {4 \pi \varepsilon_0} \frac {Q_1 Q_2} {r^2} \vec r$

The Attempt at a Solution

a) As the charges are equally spaced over that circumference, the net force acting over $+Q$ is zero, because the forcs diametrically opposed to each other cancel each other out.

b) If you take the "3'o'clock" particle out the circumference, the only particle that exerts force over $+Q$ is the particle diametrically oposed to "3'o'clock" particle: $$ \vec F= \frac {1} {4 \pi \varepsilon_0} \frac {q Q} {R^2} \vec R $$

c) The system back to equilibrium state.

What do you think guys? I'm not sure about my answers...

 

[kuruman]

In (b) you say

... the only particle that exerts force over $+Q$ is the particle diametrically oposed to "3'o'clock" particle

Have the other particles stopped exerting forces on $+Q$? Your conclusion is correct but your justification needs to be improved.

 

[Mutatis]

In (b) you say
Have the other particles stopped exerting forces on $+Q$? Your conclusion is correct but your justification needs to be improved.

They're still exerting force over $+Q$, but as they're are diametrically opposed to each other, so they cancel out. Right?

 

[kuruman]

Right. As an enrichment to this problem, what would the net force on the central charge be if you had 11 equally spaced charges on the circumference instead of 12? Your cancellation in pairs argument will no longer work. What if you removed one of these 11 and were left with 10?

 

[Mutatis]

Right. As an enrichment to this problem, what would the net force on the central charge be if you had 11 equally spaced charges on the circumference instead of 12? Your cancellation in pairs argument will no longer work. What if you removed one of these 11 and were left with 10?

In the first case, the net force is going to be a sum of the individual contributions of each charge acting over $+Q$, superposition principle. And then if I was left with 10 equally spaced charges the system is going to equilibrium state.

 

[kuruman]

In the first case, the net force is going to be a sum of the individual contributions of each charge acting over $+Q$, superposition principle. And then if I was left with 10 equally spaced charges the system is going to equilibrium state.

Sorry, you are incorrect on both counts, in fact you got it backwards. First draw the 11 force vectors acting on the central charge. Then add them using the "tail-to-tip" graphical method for adding vectors. What do you get? Also consider this. If you remove a charge from the distribution, the electric force at the center would be the same as the electric force from the old distribution plus a charge of opposite sign placed at the location of the charge that you removed. This is superposition. So if the old distribution gives zero force at the center, removing one $+q$ charge from the 12 o' clock position will provide at the center the force of a $-q$ charge placed at the 12 o' clock position.

 

[PeroK]

First draw the 11 force vectors acting on the central charge. Then add them using the "tail-to-tip" graphical method for adding vectors. What do you get?

If the arrangement is symmetrical, then the central charge cannot move in any direction, as you can rotate the system by $2 \pi/11$ and get an equivalent system where it is now moving in a different direction.

 

[Mutatis]

Sorry, you are incorrect on both counts, in fact you got it backwards. First draw the 11 force vectors acting on the central charge. Then add them using the "tail-to-tip" graphical method for adding vectors. What do you get? Also consider this. If you remove a charge from the distribution, the electric force at the center would be the same as the electric force from the old distribution plus a charge of opposite sign placed at the location of the charge that you removed. This is superposition. So if the old distribution gives zero force at the center, removing one $+q$ charge from the 12 o' clock position will provide at the center the force of a $-q$ charge placed at the 12 o' clock position.

I'll try to write up this when I get home. This exercise have got my brain confused. At the first question that I've posted, would you, if you were my physics teacher, consider it right?

 

[PeroK]

I'll try to write up this when I get home. This exercise have got my brain confused. At the first question that I've posted, would you, if you were my physics teacher, consider it right?

You need to say what that vector $\vec{R}$ means. Otherwise, it's fine.

 

[kuruman]

If the arrangement is symmetrical, then the central charge cannot move in any direction, as you can rotate the system by $2 \pi/11$ and get an equivalent system where it is now moving in a different direction.

The symmetry argument is, of course, applicable and correct in this case. More generally though, if the sum of the forces forms a closed polygon (regular or not) when drawn graphically, then that sum is zero. That's the connection that I wanted @Mutatis to establish.