Two small balls of mass 1.5 g each, with charges +0.50 µC and -0.50 µC, sit at the ends of a light…

Step 1. The moment of inertia. $I = 2ma^2 = 2 \times 1.5 \times 10^{-3}\ \text{kg} \times (0.030\ \text{m})^2 = 2.7 \times 10^{-6}\ \text{kg m}^2$.

Step 2. The torque constant. $p = q \times 2a = 0.50 \times 10^{-6}\ \text{C} \times 0.060\ \text{m} = 3.0 \times 10^{-8}\ \text{C m}$. Then $pE = 3.0 \times 10^{-8} \times 2.0 \times 10^{4}\ \text{N m} = 6.0 \times 10^{-4}\ \text{N m}$.

Step 3. The period. For small swings, $T = 2\pi\sqrt{\dfrac{I}{pE}} = 2\pi\sqrt{\dfrac{2.7 \times 10^{-6}}{6.0 \times 10^{-4}}}\ \text{s} = 2\pi \times 0.0671\ \text{s} = 0.421\ \text{s}$.

Step 4. The frequency. $f = \dfrac{1}{T} = \dfrac{1}{0.421\ \text{s}} = 2.37\ \text{Hz}$.