# Electrical Attraction

1. Aug 5, 2010

### error404

hello engineers
i have a question about electrical attraction...
i if had 2 plates, each of area 5cm2, one of them holding the charge of 3x10-4 coulomb and the other holding the same but negative charge (-3x10-4 coulomb), and the distance between the 2 plates was 1 meter, how can i find the attraction force between them?? what rule shall i use???
NOTE: the 2 plates are saperated by air.

i've been searching for a while, but nobody gave me a clear answer.
i hope u can help me out with this

thanks

2. Aug 5, 2010

### Bob S

1. Calculate capacitance:; C = ε0A/x

where ε0 = 8.85 x 10-12 Farads per meter, A = area of plates, and x is separation.

2. Calculate stored energy E = ½CV2

Q = CV

So E = ½Q2/C

3 Calculate force: F = dE/dx

Bob S

3. Aug 5, 2010

### Staff: Mentor

Or, since the plates are small compared to their separation (so you can almost treat them as point charges), you could use Coulomb's Law:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

.

4. Aug 8, 2010

### error404

thanks for replying mt Bob S, Berkeman

Bob S: i'm not sure what voltage u're talking about in the second step i didnt know what to use to replace the "V" in the stored energy rule

Berkeman: i used coulomb's law of attraction, and the result wasnt logical, the result is 8 X 10^2 which means....that if u tie 80 KGs to one capacitor, it will pull them to the other capacitor! what do u think

sorry for my belated reply
thanks

5. Aug 8, 2010

### sophiecentaur

The V would come from Q = CV, once you have calculated the capacity between the plates and then calculated the voltage needed to accumulate this charge, you may see some very big numbers.

I have frequently set my students to compare the electric force between unit charges, separated by a unit distance with the gravitational force between two unit masses, separated by a unit distance. It's just as well that both positive and negative charges exist so that they can 'almost' balance out on most occasions.

6. Aug 8, 2010

### uart

V=Q/C, but don't use the formula given in post #2 for C as it's only valid for plate separation which is small compared to the plates dimensions. It's out by several orders of magnitude in this case.

Yes, that's the correct (approx) answer. Please bare in mind that 3e-4 Coulomb is a relatively large amount of charge for plates of the given size and separation. Did you know that you'd need to apply several hundred million volts across those plates in order to get that amount of charge separation. Surprised?

Last edited: Aug 8, 2010
7. Aug 8, 2010

### sophiecentaur

Basically, '1/epsilon nought' is a very big number!
The Electric force is a very strong one c/w Gravity.

8. Aug 9, 2010

### error404

ok fine, i got that...but...can i ask u something (forget about ALL previous questions)????

if i get a capacitor with "10 micro Farad capacitace" and connect it to a battery with "30 Volts", this would give me the 3e-4 charge on each, the space between the 2 plates now is around 1 mm, if u calcualte thier attraction force, it will be huge...how come the capcitor doesnt collapse??

9. Aug 9, 2010

### sophiecentaur

There will be some force but epsilon is a lot bigger because there is a dielectric in between. I have a feeling that the majority of the stress will be internal to the dielectric.

10. Aug 9, 2010

### error404

well, if u calcualte the attraction force according to coulomb's law which is approx. true, u'll have an attraction force of 8.1X10^8, the highest relative static permettivity known according to this table is 100000, even if it was used as insulator between the 2 plates, still....u will have a force of 8.1X10^3, which is a very big number too

11. Aug 9, 2010

### sophiecentaur

OK, but coulomb's law would be between point charges (?). Perhaps the way to look at it is as follows.
Breaking each plate up into elemental points and integrating the forces between each point and each point on the other side would be a lot less; each elemental charge would only be 'near' a few elemental charges on the other plate and the inverse square law would mean that the forces between the more distant charges would be a lot less. (Charges on each plate would, of course, be separated by mutual repulsion so you'd have a uniform distribution on each plate.
I haven't done the sums but it seems reasonable that this would produce a much much smaller value of force than the one you arrived at.
Using another approach, the force would only be in the region of EQ, where E is the field (30kV/m) and the charge would be CV (= 3e-4Coulombs). That wouldn't be so much: about 10N! I could believe that the force could suspend 1kg.

12. Aug 10, 2010

### Bob S

This is a very interesting problem, and has two solutions.
We have V=30 volts, C= 10μF = εε0A/d, d=1mm, and Q = ±300 microCoulombs. (ε = relative permittivity)

First, for the record, A = 1130 square meters for an air dielectric.

1) We first consider the force for the charge Q = constant and voltage V variable.

We can write the stored energy as W = ½Q2/C

The force between the plates is F = -(∂W/∂d)Q=const = -½Q2/Cd

Putting in numbers, we get F = -4.5 Newtons. (attractive force as expected)

2) We now consider the force for the applied voltage V = constant and charge Q variable

We can write the stored energy as W = ½CV2

The force is now F = -(∂W/∂d)V=const = +½CV2/d

Putting in numbers, we now get +4.5 Newtons (repulsive force)

So the sign of the force depends on whether the charge Q or the voltage V is held constant as the separation d is varied.

Bob S

13. Aug 10, 2010

### uart

Hi Bob. You're forgetting that in the constant voltage case that power can flow to/from the voltage source as well. It's fairly easy to show (for the const voltage case) that yes the stored energy decreases in the capacitor as "d" increases, but exactly twice that amount (of lost stored energy) is transferred back to the voltage source. So in effect the amount of mechanical input power is still positive and the hence force is still attractive! The force is in fact exactly the same magnitude and direction as the previous case.

I don't have time now but can post more details later if you like.

14. Aug 10, 2010

### Bob S

Please do.

In the constant voltage case, the stored energy is W = ½CV2 = ½εε0AV2/d

In the constant charge case, the stored energy is W = ½Q2/C = ½dQ2/εε0A

In one case, the stored energy in the capacitor increases when d decreases, while in the other, the stored energy decreases when d decreases. I am fully aware that in the constant voltage case, the power supply does work. But the force between the plates depends only on how the stored energy in the capacitor depends on the plate spacing. So why is the force attractive in both cases?

Bob S

15. Aug 11, 2010

### uart

No problems.

$$F dx = dW_{mech-out} = dW_{elec-in} - dW_{stored}$$

$$dW_{stored} = \frac{1}{2} V^2 dC$$

$$dW_{elec-in} = V dQ = V^2 dC$$

Therefore,
$$F dx = V^2 dC - \frac{1}{2} V^2 dC = \frac{1}{2} V^2 dC$$

And,
$$F = \frac{1}{2}\, V^2 \, \frac{dC}{dx} = - \frac{1}{2} \, C \, V^2 / x$$

Where "x" is the plate seperation.

Note that I've derived the force in terms of output mechanical power, so that a positive value corresponds to F in the same direction as dx. The negative result therefore corresponds to an attractive force.

16. Aug 13, 2010

### error404

Bob S, uart, sophiecentaur
Thanks for helping me with this, i've been searching for the perfect answer for a really long time, now things are much more clear for me :)

Error404

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