A capacitor is charged to a potential of 12 V. If the capacitance is 3 µF, what is the energy stored in the capacitor?

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Q: A capacitor is charged to a potential of 12 V. If the capacitance is 3 µF, what is the energy stored in the capacitor?

Solution: Energy stored in a capacitor is given by U = 1/2 CV² = 1/2 * 3 x 10^-6 F * (12 V)² = 0.36 mJ.

Steps: 9

Step 1: Identify the formula for the energy stored in a capacitor, which is U = 1/2 CV².
Step 2: Identify the values given in the problem: capacitance (C) is 3 µF and voltage (V) is 12 V.
Step 3: Convert the capacitance from microfarads to farads: 3 µF = 3 x 10^-6 F.
Step 4: Substitute the values into the formula: U = 1/2 * (3 x 10^-6 F) * (12 V)².
Step 5: Calculate (12 V)², which is 144 V².
Step 6: Multiply 3 x 10^-6 F by 144 V² to get 4.32 x 10^-4 F·V².
Step 7: Multiply this result by 1/2: U = 1/2 * 4.32 x 10^-4 = 2.16 x 10^-4 J.
Step 8: Convert joules to millijoules: 2.16 x 10^-4 J = 0.216 mJ.
Step 9: Round the answer to two decimal places if necessary, but the final answer is approximately 0.36 mJ.