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

• Mutatis
In summary, the net force acting over the central charge is zero because the forcs diametrically opposed to each other cancel each other out.
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.

In (b) you say
Mutatis said:
... 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
kuruman said:
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?

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
kuruman said:
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.

Mutatis said:
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.

Mutatis and PeroK
kuruman said:
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
kuruman said:
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?

Mutatis said:
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.

PeroK said:
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.

Mutatis

## 1. What is net force acting on a charged particle?

The net force acting on a charged particle is the total force that is exerted on the particle by all other charged particles in its surrounding environment. It takes into account both the magnitude and direction of the forces.

## 2. How is net force calculated for a charged particle?

The net force on a charged particle is calculated using Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

## 3. What factors affect the net force on a charged particle?

The net force on a charged particle is affected by the magnitude of the charges of the particles involved, the distance between them, and the medium in which they are located. The presence of other nearby charged particles can also affect the net force.

## 4. What happens to the net force on a charged particle when it is in an electric field?

When a charged particle is in an electric field, it will experience a net force due to the interaction between its own charge and the electric field. The magnitude and direction of this net force will depend on the strength and direction of the electric field at that point.

## 5. Can the net force on a charged particle ever be zero?

Yes, the net force on a charged particle can be zero if it is in a state of equilibrium. This means that the forces acting on the particle are balanced, and there is no overall acceleration. For example, if a positively charged particle is placed between two equally charged particles, the net force on it will be zero.

• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
243
• Introductory Physics Homework Help
Replies
6
Views
280
• Introductory Physics Homework Help
Replies
10
Views
456
• Introductory Physics Homework Help
Replies
12
Views
417
• Introductory Physics Homework Help
Replies
3
Views
900
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
23
Views
857
• Introductory Physics Homework Help
Replies
12
Views
391