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 ($3.46×10^{-13}$), 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 $3.46×10^{-13}$, which multiplied by the sign of the formula would be $-3.46×10^{-13}$. 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.
Best Answer
$U$ 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.