Some confusion about work done to separate charges

[Jaccobtw]

Homework Statement

A carbon atom's nucleus and its electrons become separated by 20nm. How much work does it take to separate them in J?

Relevant Equations

$k \frac {q_1 q_2}{r}$

The reason I'm posting this is because I'm confused about the reasoning behind the equation. For oppositely charged particles, wouldn't work done increase with distance? According to this equation you get a higher magnitude of work done the smaller the distance. How can that be? I got the answer right, I just don't understand how this equation works. Some help is much appreciated, thank you.

 

[Herman Trivilino]

The equation you posted is the expression for potential energy, which is the work done by an external agent to assemble the charged particles. For oppositely charged particles it gives you a negative number. As you increase the distance between the particles the magnitude of the potential energy decreases, but since it's negative, that's an increase. Like when you change from $-5$ to $-3$ that's an increase, but it's a decrease in magnitude.

 

[Jaccobtw]

The equation you posted is the expression for potential energy, which is the work done by an external agent to assemble the charged particles. For oppositely charged particles it gives you a negative number. As you increase the distance between the particles the magnitude of the potential energy decreases, but since it's negative, that's an increase. Like when you change from $-5$ to $-3$ that's an increase, but it's a decrease in magnitude.

What equation would you have used?

 

[BvU]

According to this equation you get a higher magnitude of work done the smaller the distance

No you don't: the charges are of opposite sign.

What equation would you have used?

Same equation. I'm not sure what the exercise composer wants you to do: take the average distance and increase that by 20 nm ?

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[haruspex]

Relevant Equations:: $k \frac {q_1 q_2}{r}$

According to this equation you get a higher magnitude of work done the smaller the distance.

First, that's a formula, not an equation. Second, as @Mister T notes, it's the formula for the energy required to bring two charges from infinity to be distance r apart. If the charges are of the same sign, it will require more energy the closer you squeeze them.

Btw, I do not understand the question as posted. Is it the energy to take them from some unstated separation in the base state to be 20nm apart, or that required to take them from 20nm apart to infinitely separated?

 

[Jaccobtw]

First, that's a formula, not an equation. Second, as @Mister T notes, it's the formula for the energy required to bring two charges from infinity to be distance r apart. If the charges are of the same sign, it will require more energy the closer you squeeze them.

Btw, I do not understand the question as posted. Is it the energy to take them from some unstated separation in the base state to be 20nm apart, or that required to take them from 20nm apart to infinitely separated?

I guess I'm supposed to know the normal separation between a carbon atom and its nucleus

 

[BvU]

I guess I'm supposed to know the normal separation between a carbon atom and its nucleus

an atom consists of a nucleus surrounded by a cloud of electrons. The average distance of the electrons to the center for a carbon atom isn't all that easy to calculate (it is not the atomic radius. For carbon I find 0.0914 nm as atomic radius) -- and for the energy you would need the average $1\over r$, which is almost, but not exactly, the same as $1\over < r >$.
Hence our puzzledness ...

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[nasu]

You are starting with a related, but wrong formula for the question. You need to calculate the work done which is a quantity that depends on the initial and final position, so it depends on two different position. You can calculate the work as the difference in potential energy (with a minus sign) between the two points so you need to have two terms in the expression for work. $$W=-(U_2-U_1)=-(\frac{kq_1q_2}{r_2}- \frac{kq_1q_2}{r_1}) $$.
You can see that this is not the same as $$ \frac{kq_1q_2}{r_{12}} $$
where r₁₂ is the displacement between the two positions.

 

[haruspex]

I guess I'm supposed to know the normal separation between a carbon atom and its nucleus

But you stated that you got the right answer, so what did you do? Please post your working.

 

[Jaccobtw]

But you stated that you got the right answer, so what did you do? Please post your working.

I just plugged the numbers into $k \frac{q_1 q_2}{r}$. Its probably technically the wrong way to do it, but the answer is close enough.

 

[Herman Trivilino]

What equation would you have used?

Not sure, because the question is so poorly worded.

How are you interpreting the statement of the problem? Do you see the distance between the particles increasing or decreasing?

 

[haruspex]

I just plugged the numbers into $k \frac{q_1 q_2}{r}$. Its probably technically the wrong way to do it, but the answer is close enough.

That is the same as finding $k \frac{q_1 q_2}{r}-k \frac{q_1 q_2}{r'}$ if $r'=\infty$. But, at a guess, you ignored the resulting minus sign and so found $k \frac{q_1 q_2}{r'}-k \frac{q_1 q_2}{r}$, so what you calculated was the energy to move the electrons from being 20nm from the nucleus to being infinitely far from it.
That fits the question if we change "become" to "start". Is it a translation?