The capacitance of the capacitor is 6 μF, the voltage between its plates is 200 V. Determine the energy
| June 10, 2021 | education
The capacitance of the capacitor is 6 μF, the voltage between its plates is 200 V. Determine the energy of the electric field of the capacitor?
Task data: C (capacity of the selected capacitor) = 6 μF (6 * 10-6 F); U (voltage between plates) = 200 V.
To determine the required energy of the electric field of the selected capacitor, we will use the formula: W = C * U2 / 2.
Let’s calculate: W = 6 * 10-6 * 200 ^ 2/2 = 0.12 J.
Answer: The required energy of the electric field of the selected capacitor is 0.12 J.
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