The electric field strength of a 29 μF capacitor is 7000 V / m. Determine the energy of the capacitor

The electric field strength of a 29 μF capacitor is 7000 V / m. Determine the energy of the capacitor if the distance between its plates is 10mm.

Initial data: C (capacitance of the capacitor) = 29 μF (29 * 10 ^ -6 F); E (electric field strength of the capacitor) = 7000 V / m; d (distance between plates) = 10 mm (10 * 10 ^ -3 m).

We calculate the energy of the capacitor by the formula: W = 0.5 * C * U ^ 2 = 0.5 * C * (E * d) ^ 2.

Let’s make the calculation: W = 0.5 * 29 * 10 ^ -6 * (7000 * 10 * 10 ^ -3) ^ 2 = 71.05 * 10 ^ -3 J = 71.05 mJ.

Answer: The energy of the capacitor is 71.05 mJ.