Question

Энергия электрического поля заряженного конденсатора электроёмкостью 300пФ равна 15мДж Определите заряд конденсатора и напряжения между его обкладками

Answer 1

Ответ:

ответ см. на фото...............

Answer 2

Calculation of Charge and Voltage in a Charged Capacitor

To determine the charge and voltage in a charged capacitor, we can use the formula:

Energy stored in a capacitor (E) = (1/2) * C * V^2

Where: - E is the energy stored in the capacitor (given as 15 mJ) - C is the capacitance of the capacitor (given as 300 pF) - V is the voltage across the capacitor (unknown)

We can rearrange the formula to solve for V:

V = sqrt((2 * E) / C)

Let's substitute the given values into the formula and calculate the voltage:

V = sqrt((2 * 15 mJ) / 300 pF)

Converting the values to SI units: - 15 mJ = 15 * 10^-3 J - 300 pF = 300 * 10^-12 F

V = sqrt((2 * 15 * 10^-3 J) / (300 * 10^-12 F))

Simplifying the expression: V = sqrt((30 * 10^-3 J) / (300 * 10^-12 F))

V = sqrt(0.1 J / 10^-9 F)

V = sqrt(0.1 * 10^9)

V = sqrt(10^8)

V = 10^4 V

Therefore, the voltage across the capacitor is 10,000 V.

To calculate the charge in the capacitor, we can use the formula:

Q = C * V

Substituting the given values: - C = 300 pF = 300 * 10^-12 F - V = 10,000 V

Q = (300 * 10^-12 F) * (10,000 V)

Q = 3 * 10^-9 C

Therefore, the charge in the capacitor is 3 nanocoulombs.

In summary, the charge in the capacitor is 3 nC and the voltage across its plates is 10,000 V.