Questions: The figures show three different systems of point charges, all of which have positive charge. In System X, three point charges of charge Q are arranged along a straight line with a spacing of 6/2. In System Y, three point charges of charge 2Q each are arranged at the corners of an equilateral triangle of side length s. In System Z, three point charges of different charges are arranged at the corners of a right triangle, as shown in the figure. Which of the following correctly ranks the electric potential energies UX, UY, and UZ of the three systems? Take the potential energy of each system to be zero when the point charges are infinitely far from one another.

Solution Steps

The potential energy of a system of point charges is given by

$U = k \sum_{i<j} \frac{q_i q_j}{r_{ij}}$

where k is Coulomb's constant, $q_i$ and $q_j$ are the charges, and $r_{ij}$ is the distance between them.

For System X, we have three charges Q separated by distance _s_/2. The potential energy is:

$U_X = k(\frac{Q^2}{s/2} + \frac{Q^2}{s/2} + \frac{Q^2}{s}) = k(\frac{2Q^2}{s} + \frac{2Q^2}{s} + \frac{Q^2}{s}) = \frac{5kQ^2}{s}$

For System Y, we have three charges 2Q separated by distance _s_. The potential energy is:

$U_Y = k(\frac{(2Q)(2Q)}{s} + \frac{(2Q)(2Q)}{s} + \frac{(2Q)(2Q)}{s}) = k(\frac{4Q^2}{s} + \frac{4Q^2}{s} + \frac{4Q^2}{s}) = \frac{12kQ^2}{s}$

For System Z, the charges are Q, Q, and $\sqrt{2}Q$ forming a right triangle with sides $s$.

$U_Z = k(\frac{Q^2}{s} + \frac{Q(\sqrt{2}Q)}{s} + \frac{Q(\sqrt{2}Q)}{s}) = k(\frac{Q^2}{s} + \frac{\sqrt{2}Q^2}{s} + \frac{\sqrt{2}Q^2}{s}) = \frac{kQ^2}{s}(1 + 2\sqrt{2})$

Final Answer