The energy W of the electric field of a capacitor with a capacity of C = 0.015 μF is 4.7 * 10 ^ -4 J.

The energy W of the electric field of a capacitor with a capacity of C = 0.015 μF is 4.7 * 10 ^ -4 J. Determine the voltage applied to the capacitor.

Data: C – known capacitance of the capacitor (C = 0.015 μF = 15 * 10 ^ -9 F); Wк – energy el. field of the capacitor used (Wk = 4.7 * 10 ^ -4 J).

To find out the voltage applied to the capacitor used, we apply the formula: Wk = C * Ux ^ 2/2, from where we express: Ux = √ (2 * Wk / C).

Let’s perform the calculation: Ux = √ (2 * Wk / C) = √ (2 * 4.7 * 10 ^ -4 / (15 * 10 ^ -9)) ≈ 250.3 V.

Answer: A voltage of 250.3 V should have been applied to the capacitor used.