1. **State the problem:** We need to find the charge stored on either plate of a capacitor with capacitance $4 \times 10^{-6}$ F when connected to a 12 V battery. 2. **Formula used:** The charge $Q$ stored on a capacitor is given by the formula: $$Q = C \times V$$ where $C$ is the capacitance and $V$ is the voltage across the capacitor. 3. **Substitute the values:** $$Q = 4 \times 10^{-6} \times 12$$ 4. **Calculate the charge:** $$Q = 48 \times 10^{-6} = 4.8 \times 10^{-5}$$ 5. **Interpretation:** The charge stored on either plate of the capacitor is $4.8 \times 10^{-5}$ coulombs. This means when the capacitor is connected to a 12 V battery, each plate holds a charge of $4.8 \times 10^{-5}$ C.