A 20𝜇F capacitor is charged 200 V. Its stored energy is:
Correct answer: C. 0.4 J
- A. 4000 J
- B. 4 J
- C. 0.4 J
- D. 2000 J
Explanation
The energy stored in a capacitor can be calculated using the formula E = 0.5 * C * V^2. Here, C is the capacitance in farads, and V is the voltage. Given that C = 20μF (which is 20 × 10^-6 F) and V = 200 V, we can substitute these values into the formula:E = 0.5 * (20 × 10^-6) * (200^2) = 0.5 * (20 × 10^-6) * 40000 = 0.5 * 0.8 = 0.4 J.Thus, the correct answer is 0.4 J. The other options are incorrect because they either vastly overestimate or underestimate the energy based on the formula and the given values.
Last updated
About Electrostatics
Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
Practise Electrostatics
831 free Electrostatics MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Physics questions like this
Physics is on 13 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
1/C_eq = 1/C_1 + 1/C_2 + 1/C_3 + ... + 1/C_n
1 N/C equals to:
1 volt x 1 ampere is equal to:
10 V potential difference is applied across the plate of the 1 µF capacitor. What is the energy stored in the capacitor?
12 J work is to be done against an existing electric field to take a charge of 0.01C from A to B. Find the potential difference between B and A: