Electric & Gravitational Fields

[mandos]

Homework Statement

(Prior to start of questions, it informs me the relative permittivity of free space is 8.9x10^-12)

2) An alpha particle (charge 3.2x10^-19) approaches to a distance of 1.5x10^-17 from a gold nucleus (charge 1.3x10^-17). Calculate:

a) the strength of the field created by the gold nucleus at the distance of the alpha particles closest approach.


2. Homework Equations + Attempt at a solution

According the back of the book, the answer is 5.2 x 10^22 Vm^-1

Because it gave me the relative permittivity of free space, I tried using that (even though it could be for other questions, like question 1, where I needed it).

So I used F= Q1,Q1 / 4xPiex8.9x10^-12xr^2

Where I used the two charges provided and I used the distance as the radius (r).

That didn't work.

So I assumed that due to the units the back of the book gives (Vm^-1), the right equation to use would be V= Ed. But, if I rearrange is to get E .. I don't have the value of V.

But I do still think that V = Ed is barking up the right tree. But how do I find V? W = VQ? But how do I know W?

The problem is, I can't think of any other equations that are suitable to use, apart from the above two that I've discussed.

Any help would be much appreciated.

 

[Delphi51]

It is very inefficient to try finding the right formula by considering the units. Far better to get to work and memorize what each of your formulas is for. For electric field, you just have about 3 of them:
1) V = Ed which should be thought of as E = V/d
This one says, "a potential difference causes an electric field between parallel plates"

2) F = Eq which says "an electric field causes a force on a charge"

3) E = kq/r^2 which says "a charge causes an electric field around itself"
(here the k may be replaced with 1/(4*pi*epsilon) in your book)

Study all three! Then when you read your problem again and think about how you are asked what E field the charge on a nucleus causes, you will INSTANTLY know the right formula to use!

 

[thomate1]

It seems like you are on the right track with using the equation V=Ed to solve this problem. To find the electric field strength at a distance of 1.5x10^-17 from the gold nucleus, we need to find the potential difference (V) between this point and a point infinitely far away from the nucleus. This is because the potential difference between two points in an electric field is equal to the work done per unit charge in moving a test charge from one point to the other.

To find the potential difference, we can use the equation V=kQ/r, where k is the Coulomb's constant (9x10^9 Nm^2/C^2), Q is the charge of the nucleus (1.3x10^-17 C), and r is the distance from the nucleus (1.5x10^-17 m). This gives us a potential difference of 7.8x10^7 V.

Now, to find the electric field strength, we can rearrange the equation V=Ed to E=V/d, where d is the distance between the two points. In this case, d is still 1.5x10^-17 m. Plugging in the value for V that we found above, we get an electric field strength of 5.2x10^22 V/m.

It's important to note that the electric field strength at a point is equal to the force per unit charge experienced by a test charge placed at that point. So in this case, the alpha particle would experience a force of 5.2x10^22 N/C when it is at a distance of 1.5x10^-17 m from the gold nucleus.

I hope this helps clarify the concept of electric fields and how to solve problems involving them. Keep up the good work!