What is the electric potential energy stored in a capacitor of 2 µF charged to 1

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Question: What is the electric potential energy stored in a capacitor of 2 µF charged to 12 V?

Options:

Correct Answer: 0.288 mJ

Solution:

U = 1/2 * C * V² = 1/2 * 2 x 10^-6 F * (12 V)² = 0.144 mJ.

What is the electric potential energy stored in a capacitor of 2 µF charged to 12 V?

What is the electric potential energy stored in a capacitor of 2 µF charged to 12 V?

Step 1: Identify the formula for electric potential energy (U) stored in a capacitor: U = 1/2 * C * V².
Step 2: Identify the values given in the problem: C (capacitance) = 2 µF and V (voltage) = 12 V.
Step 3: Convert the capacitance from microfarads to farads: 2 µF = 2 x 10^-6 F.
Step 4: Substitute the values into the formula: U = 1/2 * (2 x 10^-6 F) * (12 V)².
Step 5: Calculate (12 V)²: (12 V)² = 144 V².
Step 6: Multiply the capacitance by the square of the voltage: 2 x 10^-6 F * 144 V² = 288 x 10^-6 J.
Step 7: Divide by 2: U = 1/2 * 288 x 10^-6 J = 144 x 10^-6 J.
Step 8: Convert joules to millijoules: 144 x 10^-6 J = 0.144 mJ.

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