A parallel plate capacitor has a capacitance of 5 µF. If the potential difference across it is increased from 10 V to 20 V, what is the change in stored energy?
Practice Questions
1 question
Q1
A parallel plate capacitor has a capacitance of 5 µF. If the potential difference across it is increased from 10 V to 20 V, what is the change in stored energy?
25 µJ
50 µJ
75 µJ
100 µJ
Energy U = 1/2 C V^2. Initial U = 1/2 * 5 × 10^-6 * 10^2 = 0.025 J; Final U = 1/2 * 5 × 10^-6 * 20^2 = 0.1 J. Change = 0.1 - 0.025 = 0.075 J = 75 µJ.
Questions & Step-by-step Solutions
1 item
Q
Q: A parallel plate capacitor has a capacitance of 5 µF. If the potential difference across it is increased from 10 V to 20 V, what is the change in stored energy?
Solution: Energy U = 1/2 C V^2. Initial U = 1/2 * 5 × 10^-6 * 10^2 = 0.025 J; Final U = 1/2 * 5 × 10^-6 * 20^2 = 0.1 J. Change = 0.1 - 0.025 = 0.075 J = 75 µJ.
Steps: 13
Step 1: Identify the formula for the energy stored in a capacitor, which is U = 1/2 C V^2.
Step 2: Note the given values: capacitance C = 5 µF (which is 5 × 10^-6 F) and initial voltage V1 = 10 V.
Step 3: Calculate the initial energy (U1) using the formula: U1 = 1/2 * C * V1^2.
Step 4: Substitute the values into the formula: U1 = 1/2 * 5 × 10^-6 * (10)^2.
