Let's solve the problem step by step. ## **Given:** - Capacitor 1: \( C_ = .05\, \mu F \) - Capacitor 2: \( C_2 = .1\, \mu F \) - Voltage: \( V = 400\, V \) - **Parallel circuit:** Voltage across each capacitor is the same. ## **Formula** The charge on a capacitor in a parallel circuit: \[ Q = CV \] --- ### **1. Charge on the first capacitor (\(.05\,\mu F\))** \[ Q_1 = C_1 \times V = .05\,\mu F \times 400\,V \] Convert \( \mu F \) to \( F \): \[ .05\,\mu F = .05 \times 10^{-6}\,F = 5 \times 10^{-8}\,F \] Calculate \( Q_1 \): \[ Q_1 = (5 \times 10^{-8}) \times 400 = 2 \times 10^{-5}\,C \] --- ### **2. Charge on the second capacitor (\(.1\,\mu F\))** \[ Q_2 = C_2 \times V = .1\,\mu F \times 400\,V \] Convert \( \mu F \) to \( F \): \[ .1\,\mu F = .1 \times 10^{-6}\,F = 1 \times 10^{-7}\,F \] Calculate \( Q_2 \): \[ Q_2 = (1 \times 10^{-7}) \times 400 = 4 \times 10^{-5}\,C \] --- ## **Final Answers:** - **Charge on the first (\(.05\,\mu F\)) capacitor:** \(\boxed{2 \times 10^{-5}\,C}\) - **Charge on the second (\(.1\,\mu F\)) capacitor:** \(\boxed{4 \times 10^{-5}\,C}\)