[Unknown008]

Qu:
The two protons of a helium nucleus are about apart. Find the electrostatic force between them. How much work must be done against this force in bringing two protons from infinity to this separation?

Ok, here's what I've done so far:

Electrostatic force = (3 sf)

Now, the second part concerns the work done, given by

which gives:



I took half the distance since the question said between the two protons. But my book says the answer is 1.5x10^-13... :(

~~~~~

Now, another one.
A capacitor is connected from two concentric spheres, the inner sphere being positively charged and the outer sphere being earthed.

Draw three sketch diagrams, using the same scales, to show how the electric field E(r) due to:
a) the charges on the inner sphere alone
b) the charges on the outer sphere alone
c) the charges of all charges
varies with distance r from the centre.

I got my first two graphs similar to the graphs of y = k/x, and both have are similar. I took both fields as positive since they are in the same direction.

For the third one, all I see is that if the radius r is the same, then the electric field will cancel out, hence, the electric field is zero. Is that a correct assumption?

[elscarta]

Regarding your first question. Notice that you answer is 8 times the correct answer. You have two 2s on the top line which don't belong there and you also need the full distance on the bottom not just half the distance (Imagine leaving one proton alone and moving the other from infinity).

[Unknown008]

You mean I don't add up the charges? Do you have the reason as to why? I simply cannot understand that clearly :(

[elscarta]

The formula for work done is the same as the formula for the force but without the r being squared on the bottom. The values of Q, Q' and r that you substituted into the first equation are the same ones that you should substitute into the second equation. After all the two charges are still the same (1.6x10^-19 each) whether you are dealing with force or work.

[Unknown008]

But don't you require more work since there is an effect of combined charges? (the two protons)

[elscarta]

Regarding your second question about the two spheres.

Let Ri be radius of inner sphere and Ro be radius of outer sphere.

Since the charge on the inner sphere is positive, this will induce an equal sized but negative charge on the outer sphere as the outer sphere is grounded.

(a) So E(r) =k/r for the inner sphere but only for r > Ri.

Inside the sphere E(r)=0. This can be explained in two ways.
Firstly at any point inside the sphere, the sum of all the electric fields due to the charges distributed evenly around the sphere = 0 (this involves calculus to prove).

Secondly consider drawing the field lines.
Electric field lines have the following properties:
- They start at positive charges and end at negative charges (or infinity).

So the field lines start at the surface of the sphere and radiate outwards in straight lines to infinity but they cannot radiate inwards as there is no negative charge there for the lines to end at and they cannot go to infinity as they are inside the sphere.

(b) Similarly E(r) = -k/r for the outer sphere but only for r > Ro. The negative value is due to the charge on the outer sphere being negative. The k value is the same as in part (a) as the charge is the same magnitude on the outer sphere as the inner sphere.

Inside the sphere E(r) = 0 for the same reasons given above.

(c) Adding the two separate fields together gives E(r) = k/r for Ri < r < Ro with E(r) = 0 inside the inner sphere and outside the outer sphere.

[elscarta]

But don't you require more work since there is an effect of combined charges? (the two protons)

The formula involves both Q and Q'. This takes care of the combined charges. Just like in the force formula.

Another way of thinking about it is that the first charge creates an electric field, the second charge then moves against the created electric field which requires work.

[Unknown008]

I got it for the second one, and thank you :)

For the first one:
Since it takes care of the combined charges, you would have twice the charge of the protons. I'm seeing it like this: first charge pushes against the proton being approached, then the second charge pushes too, thereby requiring twice the work that would instead be used if there were only one... :( I still don't get it

[elscarta]

I got it for the second one, and thank you :)

For the first one:
Since it takes care of the combined charges, you would have twice the charge of the protons. I'm seeing it like this: first charge pushes against the proton being approached, then the second charge pushes too, thereby requiring twice the work that would instead be used if there were only one... :( I still don't get it

You are confusing yourself by using the word charge and proton as if they are different things. The situation involves two protons only, each proton has a charge associated with it. You make it sound like there are three things involved, two charges and the approaching proton.

The following describes the situation:
Imagine that you have very long arms and are holding a proton in each hand at infinity, one to your left and the other to your right. In front of you there is nothing so as you bring the proton in your left hand all the way to in front of you, NO work is done since there is no force pushing back on it. Now do the same with the proton in your right hand. As it approaches the other proton in front of you, you feel a force pushing back on both hands and therefore are pushing with both your hands.

But your left hand is not moving the proton in it so no work is done by your left hand (Work = force x distance), only your right hand is doing work as it moves its proton. The formula stated calculates the work done with Q and Q' both equal to 1.6 x 10^-19 as each charge is a proton.

Hope this clears up your confusion.