Question

Определить єнергию єлектрического поля конденсатора емкостью 20 мкФ, если напряжение, приложенное к конденсатору, равно 220 В.

Answer 1

Энергия 
$W= \frac{CU ^{2} }{2}$

$W= \frac{20*10 ^{-6}*220 ^{2} }{2} =0.484 Dj$ ответ

Answer 2

Calculation of Electric Field Energy in a Capacitor

To determine the energy of an electric field in a capacitor, we can use the formula:

E = (1/2) * C * V^2

Where: - E is the energy of the electric field in the capacitor, - C is the capacitance of the capacitor, and - V is the voltage applied to the capacitor.

In this case, the capacitance of the capacitor is given as 20 μF (microfarads), and the voltage applied is 220 V.

Substituting these values into the formula, we get:

E = (1/2) * 20 μF * (220 V)^2

Now, let's calculate the energy of the electric field in the capacitor.

E = (1/2) * 20 μF * (220 V)^2

Using the given values, we can calculate the energy as follows:

E = (1/2) * 20 * 10^(-6) F * (220 V)^2

E = (1/2) * 20 * 10^(-6) F * 48400 V^2

E = 0.01 J

Therefore, the energy of the electric field in the capacitor with a capacitance of 20 μF and a voltage of 220 V is 0.01 J.

Please note that the calculation is based on the given values and the formula for the energy of an electric field in a capacitor.