# [Physics] Need help with signs (+ -) in electric potential work and potential energy problem

conventionselectrostaticspotential energywork

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.

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