I'm lost in physics. horribly lost. We're studying electric potential and work. totally lost. the notes she gave us do not include anything about distance, yet our worksheets involves distance multiple times. so i'm just lost. i will post one of my quesitons, and i'd love some explanation...

Two parallel conducting plates are connected
to a constant voltage source. The magnitude
of the electric field between the plates is
2107 N

Answer in units of N/C.

You have to assume that the E field is uniform throughout the region where the plates are (because... figure it out yourself. I give you a tip: what is the E field of an infinite charged plate?). With this assumption, E=V/d. You can then solve easily the problem.

Why does E=V/d? It follows from the definition of the potential function, namely E=-grad(V). Thus V(r)=-int _(infinity)^r E dr where E and dr are vectors. It turns out that if E is parallel to dr, such as in this case, and that E is constant, the simplification done holds.

__________________
Isaac
If the problem is too hard just let the Universe solve it.

Hi there,

There are two ways you can solve this problem and actually come up with a value. The first is basically what Arbolis had to say about the constant E field between the plates of the capacitor and using the formula V = E"dot"d = Edcos(theta) where theta is the angle between the E field and d is the separation distance between the plates that points in the same direction as the E field. Since they both point in the same direction, theta is zero and the cosine of zero is 1, so V = Ed is the final equation.

Using this equation you can first multiply the voltage by 4 and make the distance 1/8 of what it was and solve for the E field that would exist if those conditions are met.

V = Ed
V = 2107d
So,
4V = V(new) = E(new)*d(new) where the "news" are the new values for that variable. Using this and placing in the new distance we get:

4V = E(new)(1/8)d where (1/8)d = d(new)
32V = E(new)d = V(new) and E(new) = V(new)/d(new) = A*(V/d) where V/d is the old E field of 2107 N/C and A is the factor by which the old field has changed by to create the new field.

So,

32V = E(new)d
32*(V/d) = E(new) and V/d = E(old) = 2107 N/C, so

E(new) = 32(V/d) = 32*2107N/C = 67424 N/C

The other way to solve this problem is to first find the new E field due to the distance change and then place that into the V = Ed formula using 4 times the old voltage and the new distance. This too will allow us to solve for the new E field. Let's start by keeping everything but one variable at a time at what it was and replace them with their new values until everything has been changed. First let's find the new E field due to the plates being brought closer together by a factor of 8.

E(old) = kq/r^2
E(new) = kq/(r/8)^2

E(new) = 64*kq/r^2 = 64*E(old)

Alright, lets place this and the new distance into the equation for the voltage across a capacitor due to the electric field that we found above, namely V = Ed; but for the time we will hold V at its initial value, it's not correct, but we will change it so it will be.

V = 64*E*(1/8)*d = 8Ed, but we still have the same voltage as before on the left hand side, all we've done so far is replace the E field value and the d value with the new values, but the new voltage is supposed to be 4 times the old voltage, so let's multiply each side by 4 so as to increase the voltage by a factor of 4.

V = 8Ed
4V = 32Ed = V(new)

E(new) = V(new)/d = 32*E(old) = 67424 N/C.

Now does this make sense, it's the same answer we got using the easier method? Let's see. If we put in for d, 1/8 of d, which should be the separation distance between the plates, we should just see an increase in voltage by a factor of 4, since that increase was independent of the change in separation distance.

V = 32E(1/8)d = 4Ed, yup four times the old voltage is the new voltage between the plates separated by 1/8 the previous distance.

So let’s say the initial distance separating the plates was 1 meter, then our new equation which is "tailored" to a separation distance that is 1/8 the initial distance can't be used with 1 meter, but it can be used with 1/8 of a meter, the voltage obtained from that distance should be 4 times the previous voltage, as stated in the initial problem.

V = 32E(1/8) = 4E

And initially we had V = Ed, so at 1 meter V = E*1 = E = (1/4)V(new), so indeed our new voltage is four times the initial voltage, which was one of the requirements at 1/8 the separation distance.

Basically what it all shows is that the new E field is 32 times stronger than the initial E field, which I believe was your question.

Many Smiles,
Craig