# Work that must be done to charge a spherical shell

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1. Feb 11, 2016

### ElPimiento

1. Calculate the work that must be done on charges brought from infinity to charge a spherical shell of radius
R = 0.100 m to a total charge of Q = 125 μC.

2. $$V = k_e\int{\frac{dq}{r}}$$$$\triangle V = - \int{E \cdot ds}$$$$W = q\triangle V$$

3. I started with assuming the spherical shell produces an electric field equal to that of a point charge, so
$$E = k_e \frac{q}{r^2}$$
$$V = k_e \frac{q}{r}$$ (since they're coming from infinity the initial potential is 0)
But once I get to this point I don't know where to go, I tried sorta just using the fundamental charge for q in
$$W = q\triangle V$$
to no success. I also tried a similar method to the aforementioned, where I started by assuming each infinitesimal bit of work could be given by:
$$dW = k_e \frac {e}{r} dq$$
But I don't know how to evaluate this as an integral.
So how should I set up this problem?

2. Feb 11, 2016

### BvU

Hello Capsicum annuum ,

So you start from an uncharged sphere and bring a little charge $dq$ from infinity to the surface of the sphere. For free, because no force is needed. We can ignore subtle complicated effects because $dq$ is infinitesimally small.

Once there is some charge q on the sphere (which -- as I hope you know -- distributes itself evenly over the outer surface), the potential at the surface is $k_e q\over R$. So then the work to bring a little charge $dq$ from infinity to the surface of the sphere is, as you wrote, more or less, $$dW = {k_e q\over R}\; dq$$
For the last bit of charge $dq$ to be added, there is (almost completely) a charge $Q$ on the sphere and you can expect that the work needed is $$dW = {k_e Q\over R}\; dq$$
Do you now see what the integral $$W = \int dW$$ should be ?

3. Feb 11, 2016

### ElPimiento

Thank you so much! That makes a lot of sense, so the field from the spherical shell is
$$W = {k_e q\over R}\;$$
so to move a charge, dq, to its surface takes an amount of work
$$dW = {k_e q\over R}\; dq$$
and R is constant in this case (obviously, I should have realized that lol) so we can move it and Coulomb's constant out of the integral:
$$W = {k_e \over R} \int (q) dq$$
evaluated from 0 to Q (idk how to put the endpoints), in order to get the total work.

Thank yo so much, his gives the right answer! But is my logic behind it right?

4. Feb 11, 2016

### BvU

Well done.

Latex uses _ for the lower limit and ^ for the upper, so \int_0^Q gives $$\int_0^Q$$ when you use the double \$ for displayed equations and $\int_0^Q$ when you use the double # for in-text equations

5. Feb 11, 2016

### gracy

@ElPimiento Take his advice on latex. That really helps.

6. Feb 11, 2016

### ElPimiento

Thanks for all the help.

7. Feb 11, 2016

### haruspex

Not quite. Your first equation, W= etc., you describe as the field. You mean the potential, and it should therefore be V=....
To get to the second equation, you multiplied the right hand side by dq, so you should do the same to the left, giving Vdq. Now you can validly replace Vdq by dW.