Q.(a) In a quark model of elementary particles, a neutron is made of one up quark [charge $\tfrac{2}{3}e$] and two down quarks [charges $-\tfrac{1}{3}e$]. Assume that they have a triangle configuration with side length of the order of $10^{-15}\ \text{m}$. Calculate electrostatic potential energy of neutron and compare it with its mass $939\ \text{MeV}$.

Concept: Inverse Square Law Comparison — electrostatic potential energy between point charges is $U = \frac{1}{4\pi\varepsilon_0} \frac{q_i q_j}{r}$, summed over all pairs.

(a) Neutron

Quarks: one up ($+\frac{2}{3}e$), two down ($-\frac{1}{3}e$). Side length $r \approx 10^{-15}\ \text{m}$.

Three pairs:

• Up–down (two such pairs): $U_1 = 2 \times \frac{1}{4\pi\varepsilon_0} \frac{(+\frac{2}{3}e)(-\frac{1}{3}e)}{r}$
• Down–down: $U_2 = \frac{1}{4\pi\varepsilon_0} \frac{(-\frac{1}{3}e)(-\frac{1}{3}e)}{r}$

Total:

$U = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r} \left[ 2\left(-\frac{2}{9}\right) + \frac{1}{9} \right] = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r} \left(-\frac{3}{9}\right) = -\frac{1}{3} \cdot \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}$

Using $\frac{1}{4\pi\varepsilon_0} \frac{e^2}{1\ \text{fm}} \approx 1.44\ \text{MeV}$ and $r = 1\ \text{fm} = 10^{-15}\ \text{m}$:

$U = -\frac{1}{3} \times 1.44\ \text{MeV} \approx -0.48\ \text{MeV}$

Neutron mass $= 939\ \text{MeV}$. Ratio $|U|/m \approx 5 \times 10^{-4}$ — electrostatic energy is negligible compared to mass.

(b) Proton

Quarks: two up ($+\frac{2}{3}e$), one down ($-\frac{1}{3}e$).

Pairs: …

The three quarks sit at the vertices of an equilateral triangle of side $r=10^{-15}\ \text{m}$. The electrostatic potential energy is the sum of the three pairwise Coulomb energies:

1. Base Coulomb energy for a pair of charges $e$ at separation $r$.

Each pair energy is this number times the product of the two charge fractions.

2. Neutron: quarks $+\tfrac{2}{3}e,\ -\tfrac{1}{3}e,\ -\tfrac{1}{3}e$.

• $u$–$d_1:\ \left(+\tfrac{2}{3}\right)\left(-\tfrac{1}{3}\right)=-\tfrac{2}{9}$
• $u$–$d_2:\ -\tfrac{2}{9}$
• $d_1$–$d_2:\ \left(-\tfrac{1}{3}\right)\left(-\tfrac{1}{3}\right)=+\tfrac{1}{9}$

Sum of the charge-fraction products $=-\tfrac{2}{9}-\tfrac{2}{9}+\tfrac{1}{9}=-\tfrac{1}{3}$, so

3. Proton: quarks $+\tfrac{2}{3}e,\ +\tfrac{2}{3}e,\ -\tfrac{1}{3}e$.

• $u_1$–$u_2:\ \left(+\tfrac{2}{3}\right)\left(+\tfrac{2}{3}\right)=+\tfrac{4}{9}$
• $u_1$–$d:\ -\tfrac{2}{9}$ …

Method: Summing Pairwise Coulomb Potential Energies for a System of Point Charges

Use this whenever you're asked for the total electrostatic potential energy of a small cluster of point charges arranged in a known geometry (and often to compare it against some other energy scale, like a rest mass).

Steps

Step 1: List every distinct pair of charges and the distance between them

Electrostatic PE is defined pairwise — there is no "PE of a single charge" in isolation. For $N$ charges there are $\binom{N}{2}$ pairs. Draw or imagine the geometry (here, an equilateral triangle) and read off the separation for each pair; if the configuration is symmetric, many pairs will share the same distance.

Step 2: Write the total PE as a sum over pairs

Each term keeps its sign — opposite-sign pairs contribute negative PE (bound, attractive), same-sign pairs contribute positive PE (repulsive).

Step 3: Factor out the common scale, then sum the dimensionless charge fractions …