2020 A/L Physics Past Paper · Question 25 (English Medium)

Answer: (1) $\dfrac{5q^2}{4\pi\varepsilon_0 a}$

What the question asks

We must find: the total electric potential energy $U$ of a system of three charges $+4q, +3q, -q$ at the corners of an equilateral triangle of side $a$.

Relevant theory

This work gets accumulated in q, as potential energy and since this was made possible due to the contribution of charge q,, this energy is considered as the potential energy of the pair q,, 7, of the charges.

The total potential energy of a system consisting of a number of charges would be the sum of the potential energies of all positive pairs of charges in the system.

Applying it to this question

Contributing pairs: three charges give three pairs — $(+4q, +3q)$, $(+4q, -q)$, $(+3q, -q)$. Since the triangle is equilateral, every pair is separated by $r = a$.

Reasoning and calculation

Pair $(+4q, +3q)$

$U_{12} = +\frac{12q^2}{4\pi\varepsilon_0 a}$

Product $(4q)(3q) = 12q^2$; like charges repel, so positive work is needed to bring them together: $U_{12} = \dfrac{12q^2}{4\pi\varepsilon_0 a}$.

Pair $(+4q, -q)$

$U_{13} = -\frac{4q^2}{4\pi\varepsilon_0 a}$

Product $(4q)(-q) = -4q^2$; opposite charges attract, so the work is negative: $U_{13} = -\dfrac{4q^2}{4\pi\varepsilon_0 a}$.

Pair $(+3q, -q)$

$U_{23} = -\frac{3q^2}{4\pi\varepsilon_0 a}$

Product $(3q)(-q) = -3q^2$; another attracting pair: $U_{23} = -\dfrac{3q^2}{4\pi\varepsilon_0 a}$.

Signs: potential energy is a scalar — no direction, but it does have a sign. A like-charge pair has $q_1q_2 > 0$ (positive); an opposite-charge pair has $q_1q_2 < 0$ (negative). So the three must be added algebraically, signs included.

$U = U_{12} + U_{13} + U_{23}$

$U = \frac{1}{4\pi\varepsilon_0 a}\left[(4q)(3q) + (4q)(-q) + (3q)(-q)\right]$

$U = \frac{(12 - 4 - 3)q^2}{4\pi\varepsilon_0 a}$

$U = \frac{5q^2}{4\pi\varepsilon_0 a}$

Result: $U = \dfrac{5q^2}{4\pi\varepsilon_0 a}$ — a positive value. The like-charge pair's $+12$ outweighs the two attracting pairs' $-7$.

Reading the options

Comparing with the options: $\dfrac{5q^2}{4\pi\varepsilon_0 a}$ — option (1).

Why the other options fail

Adds the charges instead of multiplying them in pairs ($4 + 3 - 1 = 6$).

Takes only the magnitude of the work to bring in $-q$; the $(+4q, +3q)$ pair is missing, and the sign is dropped.

The pair $(+3q, -q)$ is missing: $12 - 4 = 8$.

Ignores the signs: $12 + 4 + 3 = 19$.

Checks

Symmetry check: all three sides are $a$, so every pair shares the same factor $\dfrac{1}{4\pi\varepsilon_0 a}$; the answer depends only on the sum of the pairwise products of the charges. Swapping which corner holds which charge would not change the answer.

Dimension check: $\dfrac{q^2}{4\pi\varepsilon_0 a}$ has unit $\dfrac{\text{C}^2}{\text{C}^2\text{N}^{-1}\text{m}^{-2} \cdot \text{m}} = \text{N m} = \text{J}$ ✓ — a unit of energy.

Second way: bring in the charges one at a time — $+3q$ costs $12$, then $-q$ costs $-4 - 3 = -7$; total $5$ ✓.

$q1 * q2 + q1 * q3 + q2 * q3$

Conclusion

$U = \frac{5q^2}{4\pi\varepsilon_0 a}$

$U = \dfrac{1}{4\pi\varepsilon_0 a}(12 - 4 - 3)q^2 = \dfrac{5q^2}{4\pi\varepsilon_0 a}$ — the answer is (1).

Treating potential energy as a vector, or dropping the signs when adding.

Potential energy is a scalar; add each pair's $q_1q_2$ with its own sign.