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The problem is from Sears and Zemansky's University Physics, 12th edition in Spanish.

The problem says: How much work is needed to assemble an atomic nucleus that contains three protons if it is shaped like an equilateral triangle, whose side is $2.00 x 10^{-15}m$ long and has a proton on each vertex? Suppose The protons started all separate from each other.

What I did was use this formula to calculate the system's electric potential when the particle were all close to each other: $$ U = \dfrac{1}{4\pi\epsilon_0} \sum_{i<j}^{} \dfrac{q_iq_j}{r_{ij}}$$ where:
$U=$ electric potential
$q_i$ and $q_j=$ pair of punctual charges (in this case the protons)
$r_{ij}=$ distance between protons

It turns out I got the numerical value right right ($3.46x10^{-13}$), according to the solutions appendix, but I didn't get the same sign as the solution. I got a negative value when the text says it is positive. I can argue the value being negative since work is defined the following way:$$W_{A{\rightarrow}B}=-{\Delta}U=U_A-U_B$$

Where:
$W_{A{\rightarrow}B}=$ work done by a force from point $A$ to point $B$
$U_A$ and $U_B=$ electric potential in points $A$ and $B$, respectively

So if we choose the starting point for all the particles to be somewhere where they are all infinitely separated, $U_A$ would be zero and $U_B$ would be $3.46x10^{-13}$, which multiplied by the sign of the formula would be $-3.46x10^{-13}$. This result is consistent with the principle that says "the work is negative whenever the electric potential increases", and the one that states "the electric potential increases whenever a charge moves in the opposite direction of the force applied to it by the electric field of another charge".

My question would be whether my answer is wrong and I am missing a concept, or if it is a textbook error.

Qmechanic
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Pobs97
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3 Answers3

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I don't think it matters really, it's a matter of convention. In your picture if we start at infinity: $$ W = \int_\infty^r dr' F(r')= C \int_\infty^r \frac{1}{r'^2} = \frac{-C}{r} \\ E_{pot} = -Q\int_\infty^r dr' E(r') = \frac{C}{r} $$ Energy conservation holds:$E_{pot}+W=0.$ Depending upon how you define work i.e. if the the system "carries out the work" or if "it absorbs it", both signs should be fine.

dodge
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$U$ is the electrostatic potential energy of the system of the three protons.

To assemble the three protons coming from infinity external work needs to be done so the electrostatic potential energy is positive.
So the textbook answer has computed the external work which needs to be done to bring the protons together.

Another approach is to compute the work done by the electric field which will be negative as you have found.

Farcher
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This result is consistent with the principle that says "the work is negative whenever the electric potential increases", and the one that states "the electric potential increases whenever a charge moves in the opposite direction of the force applied to it by the electric field of another charge".

My question would be whether my answer is wrong and I am missing a concept, or if it is a textbook error.

I think one is trying to answer the question :

How much work is needed to assemble an atomic nucleus that contains three protons if it is shaped like an equilateral triangle, whose side is 2.00x10−15m long and has a proton on each vertex? Suppose The protons started all separate from each other.

Therefore what we would do is to assemble the above charges and naturally a positive amount of 'work' will be needed to assemble them.

As any agency bringing a proton from infinity to the one of the vortex of triangular arrangement will need to do work against the coulomb repulsion.

therefore one should not get confused by such statements like.... work done is always negative when potential energy increases....etc.

As potential energy in a force field depends on the nature of charges as well as their relative position in the field.

The work done and its transformation as potential energy of a system depends on whether the external agency is doing the work or the field itself is putting in work to make an arrangement of bodies in a field.

My conclusion is that the work needed to assemble the three protons should be positive .

A prime example would be to asses the work done to assemble a proton and an electron say about one Angstrom apart. Initially both particles are at infinite distances from each other.

The field due to proton extends up to infinity. The movement of electron from infinity to the designated place near proton will not need any external agency to do work -

Rather the field itself will do the necessary work on the electron to bring it at the place - naturally the work needed from any external agent will not be positive. The potential energy of this system will be negative and its a bound system. Whereas our three protons are not bound but can fly apart and has "potential" to do do work.

The two principles given by author (quoted above)are not consistent with each other and that's why a negative sign is coming up in the answer.

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drvrm
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