Question

Определите энергию электрического поля конденсатора, если заряд одной из его пластин равен 0,2 мКл, а напряжение между обкладками равно 300В. Помогите срочно!!!

Answer 1

Ответ:

Задача в одну формулу,надо только нанокулоны перевести.

W=q*U/2=20*10^(-9)*300/2=3*10^(-6) Дж

Answer 2

Calculation of the Electric Field Energy of a Capacitor

To calculate the electric field energy of a capacitor, we need to know the charge on one of its plates and the voltage between the plates.

Given: - Charge on one of the plates (q) = 0.2 mC (milliCoulombs) - Voltage between the plates (V) = 300 V

The electric field energy (U) of a capacitor can be calculated using the formula:

U = (1/2) * q * V

Substituting the given values into the formula, we get:

U = (1/2) * 0.2 mC * 300 V

To perform the calculation, we need to convert the charge from milliCoulombs to Coulombs:

U = (1/2) * 0.2 * 10^-3 C * 300 V

Simplifying the expression, we find:

U = 0.03 J (Joules)

Therefore, the energy of the electric field in the capacitor is 0.03 Joules.

Please note that the above calculation assumes an ideal capacitor with no losses or other factors that may affect the energy stored in the electric field.