Physics help

[trigkid213]

I need help on physics homework. We submit most of our homework online (it's called CAPA). I'm on the last problem:

The plates of a 580 pF capacitor are seperated by a distance of 7.21 mm. The space is filled with air. The capacitor is connected to a 12 V source.
A. What is the electric field between the plates?
B. How much force pulls the plates together?

Can some give me the formula for electric field strength in relation to distance between the plates? I need it in the next ten or fifteen minutes.

Edit: Gotta go.

[todlangweilig]

b.

Coulomb Force Law - the magnitude of the force of interaction between two point charges is proportional to the product of the charges and inversely proportional to the distance squared between the two charges, ie

F = k( q1* q2) / r^2 where k = 9x10^9 Nm^2/C^2


and the direction of the force is along the line connecting the two charges

hope that what u need

[Raster]

I should know this for being such a tech geek... but oddly enough, I dont... I guess I am more of a digital guy :biggrin: .

[Arcane Wizard]

I've had this in highschool, but question a is rather strange for me "what is the electric field", the electric field is the electric field???? Or maybe the electric field is nice?? a particular colour? political association?

??

[trigkid213]

I meant the strength of the electric field, which is a ratio of force to charge (in Newtons per Coulomb, or N/C).

[DigiTan]

You could probably use the E = Voltage/distance formula to get E = 1660 Volts/meter. That sounds like a ridiculously high, but its about right.


That one came from Gauss's Law (∫ E ∙ dA = ρfree/ε). That basically says, "the total electric field (E) from a closed surface made up of infinately small sections (dA) is equal to the total charge inside (ρfree) divided by permittivity of the material it's in (ε).

Anyway, because it's dealing with a surface area, you'd want to introduce area into the equation. ρfree is the same as "area * ρfree/area," so you call "ρfree/area" σ (density), and call "area" A...

∫ E ∙ dA = σ A /ε

This part gets kind of hard to visualize. We assume the plate is completely flat and we measure the total E field coming out of it drawing a boundary around it and using the integral to "count" the total E entering and/or leaving the boundary. To simplify the math as much as possible, the boundary is usually shaped like a flattened cylinder. Since the flattened walls have almost 0 height, we ignore them and concentrait on only the top and bottom. So the formula becomes.

∫ dA + ∫ dA + 0 = σ A /ε
^top ^bottom ^ walls

Regardless of the coordinate system, ∫ dA = A always. Now that the integrals are gone, you solve for E...

E A + E A + 0 = σ A /ε
2EA = σ A /ε
2EA / A = σ /ε
2 E = σ /ε
2 E / 2 = σ / 2 ε
E = σ / 2 ε

...So now "E" is the electric field vector coming out of 1 plate. We know the E field always points toward negative charges and away from positive charges. Treat the fields from the plates as separate fields. One field (-) will be point toward its plate, the other (+) will be pointing away from its plate. To two fields will add up to become doubly strong, so the field in between is...

Etotal = 2*E
Etotal = σ / ε

So it's really the charge per area, divided by the permittivity of the material. Using Ampere's and Faraday's laws, it's possible to link this to the E = V/d formula, but that's a huge pain.

[trigkid213]

Another question:

Protons move in a circle of radius 5.60 cm in a 0.548 T magnetic field. What value of electric field could make their paths straight?