Capacitors
A parallel-plate capacitor of capacitance 5 μF is charged to a potential of 200 V. After charging, the battery is disconnected. What is the energy stored in the capacitor?
Select the correct option:
Solution
0.1 J
The energy stored in a capacitor is given by (U = \frac{1}{2}CV^2), which can also be written as (U = \frac{Q^2}{2C} = \frac{1}{2}QV). All three forms are equivalent; (U = \frac{1}{2}CV^2) is most convenient when capacitance and voltage are given. The energy is stored in the electric field between the plates, with energy density (u = \frac{1}{2}\varepsilon_0 E^2). Here, (C = 5 \times 10^{-6}) F and (V = 200) V. Substituting: (U = \frac{1}{2} \times 5 \times 10^{-6} \times (200)^2 = \frac{1}{2} \times 5 \times 10^{-6} \times 4 \times 10^4 = \frac{1}{2} \times 0.2 = 0.1) J. Option 0.05 J is incorrect because it omits the factor of ((200)^2) and uses V instead of (V^2). Option 0.2 J is incorrect because it drops the factor of (1/2). Option 0.025 J is incorrect because it uses (\frac{1}{4}CV^2) instead of (\frac{1}{2}CV^2). The disconnect after charging ensures the charge remains constant. Plausibility check: 0.1 J for a 5 μF capacitor at 200 V is reasonable — a large capacitor at high voltage stores measurable energy.
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About This Question
- Subject
- physics
- Chapter
- electrostatics
- Topic
- capacitors
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
0.1 J
The energy stored in a capacitor is given by (U = \frac{1}{2}CV^2), which can also be written as (U = \frac{Q^2}{2C} = \frac{1}{2}QV). All three forms are equivalent; (U = \frac{1}{2}CV^2) is most convenient when capacitance and voltage are given. The energy is stored in the electric field between the plates, with energy density (u = \frac{1}{2}\varepsilon_0 E^2). Here, (C = 5 \times 10^{-6}) F and (V = 200) V. Substituting: (U = \frac{1}{2} \times 5 \times 10^{-6} \times (200)^2 = \frac{1}{2} \times 5 \times 10^{-6} \times 4 \times 10^4 = \frac{1}{2} \times 0.2 = 0.1) J. Option 0.05 J is incorrect because it omits the factor of ((200)^2) and uses V instead of (V^2). Option 0.2 J is incorrect because it drops the factor of (1/2). Option 0.025 J is incorrect because it uses (\frac{1}{4}CV^2) instead of (\frac{1}{2}CV^2). The disconnect after charging ensures the charge remains constant. Plausibility check: 0.1 J for a 5 μF capacitor at 200 V is reasonable — a large capacitor at high voltage stores measurable energy.
This medium difficulty physics question is from the chapter electrostatics, covering the topic of capacitors. It appeared in the 2025 exam.
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