Four point charges q₁ = 1 $\mu$C, q₂ = 2 $\mu$C, q₃ = 3 $\mu$C, and q₄ = 4 $\mu$C are located at the corners of a square whose side s = 1.00 m. The electrostatic potential energy of this system of charges is

Question

Quizplus · Accepted Answer

The electrostatic potential energy (U) of a system of point charges is given by the formula:

$U=\frac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{q_iq_j}{4\pi\epsilon_0r_{ij}}$

where $q_i$ and $q_j$ are the magnitudes of the point charges, $\epsilon_0$ is the permittivity of free space, and $r_{ij}$ is the distance between the charges.

For the given system of charges, the distance between any two charges is $s\sqrt{2}$ (diagonal of the square). Using this distance and the given charges, we can calculate the potential energy of the system:

$U=\frac{1}{2}\left(\frac{1\cdot2}{4\pi\epsilon_0(s\sqrt{2})}+\frac{1\cdot3}{4\pi\epsilon_0(s\sqrt{2})}+\frac{1\cdot4}{4\pi\epsilon_0(s\sqrt{2})}+\frac{2\cdot3}{4\pi\epsilon_0(s\sqrt{2})}+\frac{2\cdot4}{4\pi\epsilon_0(s\sqrt{2})}+\frac{3\cdot4}{4\pi\epsilon_0(s\sqrt{2})}ight)$

Simplifying this expression, we get:

$U=\frac{7}{8\pi\epsilon_0s}$

Substituting the given values, we get:

$U=0.286	ext{ J}$

Therefore, the best choice is (C), which is 0.286 J.

Four Point Charges Q1 = 1 μ\muμ C, Q2 = 2 μ\muμ

Four Point Charges Q₁ = 1 $\mu$ C, Q₂ = 2 $\mu$