Question

Площадь пластины конденсатора 36квадратных сантиметров тольщина слоя диэлектрика 0,14 сантиметра вычислить электро емкость ,заряд и энергию запасенную конденсатором если разность потенциалов на его пластинах 300вольт ,а диэлектрическая проницаемость слюды равна 7

Answer 1

Емкость будет 18*Eo
считается по формуле (Ео*Е*S)/d
чему равна эпсилон нулевое я не помню )
напряженность 300/0,14=2142,8571 В/см

Answer 2

Calculation of Capacitance, Charge, and Energy Stored in a Capacitor

To calculate the capacitance, charge, and energy stored in a capacitor, we need to use the formula:

C = (ε₀ * εᵣ * A) / d

Where: - C is the capacitance of the capacitor, - ε₀ is the vacuum permittivity (8.854 x 10⁻¹² F/m), - εᵣ is the relative permittivity (dielectric constant) of the material between the capacitor plates, - A is the area of the capacitor plates, and - d is the distance (thickness) of the dielectric layer.

Given the following values: - Area of the capacitor plates (A) = 36 square centimeters = 36 cm², - Thickness of the dielectric layer (d) = 0.14 centimeters = 0.14 cm, - Potential difference across the capacitor plates (V) = 300 volts, and - Dielectric constant of mica (εᵣ) = 7.

Let's calculate the capacitance, charge, and energy stored in the capacitor.

Calculation:

1. Convert the area and thickness to square meters and meters, respectively: - Area of the capacitor plates (A) = 36 cm² = 36 * 10⁻⁴ m² - Thickness of the dielectric layer (d) = 0.14 cm = 0.14 * 10⁻² m

2. Substitute the given values into the capacitance formula: - C = (ε₀ * εᵣ * A) / d - C = (8.854 x 10⁻¹² F/m * 7 * 36 * 10⁻⁴ m²) / (0.14 * 10⁻² m)

3. Calculate the capacitance (C): - C = 0.177 F

4. Calculate the charge (Q) stored in the capacitor: - Q = C * V - Q = 0.177 F * 300 V

5. Calculate the energy (E) stored in the capacitor: - E = (1/2) * C * V² - E = (1/2) * 0.177 F * (300 V)²

Results:

Based on the calculations, the values are as follows: - Capacitance (C) = 0.177 Farads (F) - Charge (Q) stored in the capacitor = 53.1 Coulombs (C) - Energy (E) stored in the capacitor = 2,389.5 Joules (J)

Please note that these calculations assume ideal conditions and do not account for factors such as leakage or losses in the capacitor.