The plates of a parallel-plate capacitor are connected to a source

The voltage of the source connected to the plates of a parallel-plate capacitor is 2 V. If the capacitor is half filled with a dielectric with a dielectric constant of 2, then it is necessary to determine the change in the energy of the electric field in the capacitor. The boundary between the dielectric and air is perpendicular to the plates, the distance between which is d = 1 cm, and the area of ​​the plates is S = 50 cm^2.

To solve the problem, it is necessary to use the formula for calculating the capacitance of a flat capacitor, which is expressed as follows: C = εS / d, where C is the capacitance of the capacitor, ε is the dielectric constant of the dielectric, S is the area of ​​the capacitor plates, d is the distance between the plates.

Based on this formula, it is possible to determine the capacitance of the capacitor, provided that it is half filled with dielectric: C' = 2εS / d.

The change in the energy of the electric field of a capacitor when filling it with a dielectric is determined by the formula: ΔW = (1/2)C'U^2 - (1/2)CU^2, where U is the voltage on the capacitor before filling with a dielectric.

Substituting known values, we get: C' = 2250 / 1 = 200 pF, U = 2 V. Then ΔW = (1/2)200(2^2) - (1/2)100(2^2) = 200 µJ.

Thus, the change in the energy of the electric field of the capacitor when filling it with a dielectric is 200 μJ.

Plates of a parallel plate capacitor

We present to your attention a digital product - “Plates of a flat-plate capacitor”.

This item contains a detailed description of a parallel plate capacitor consisting of two plates connected to a voltage source.

The capacitor has the following features:

• The distance between the plates is 1 cm
• The area of ​​the plates is 50 cm²
• The voltage source is connected to the plates with a voltage of 2 V
• It is possible to fill the capacitor halfway with a dielectric with a dielectric constant of 2

This product will be useful for those interested in electronics and physics, as well as for students studying electrical circuits and capacitors.

Don't miss your opportunity to purchase this digital item today!

This product is a description of a parallel plate capacitor consisting of two plates connected to a voltage source with a voltage of 2 V. The distance between the plates is 1 cm, and the area of ​​the plates is 50 cm². The description also contains information about the possibility of half filling the capacitor with a dielectric with a dielectric constant of 2 and the boundary between the dielectric and air located perpendicular to the plates.

To solve the problem, it is necessary to use the formula for calculating the capacitance of a flat capacitor, which is expressed as follows: C = εS / d, where C is the capacitance of the capacitor, ε is the dielectric constant of the dielectric, S is the area of ​​the capacitor plates, d is the distance between the plates. Based on this formula, it is possible to determine the capacitance of the capacitor, provided that it is half filled with dielectric: C' = 2εS / d.

The change in the energy of the electric field of a capacitor when filling it with a dielectric is determined by the formula: ΔW = (1/2)C'U^2 - (1/2)CU^2, where U is the voltage on the capacitor before filling with a dielectric.

Substituting the known values, we get: C' = 2250 / 1 = 200 pF, U = 2 V. Then ΔW = (1/2)200(2^2) - (1/2)100(2^2) = 200 μJ .

Thus, the change in the energy of the electric field of the capacitor when filling it with a dielectric is 200 μJ.

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The subject of the description is a flat capacitor whose plates are connected to a source with an emf. 2 V. The capacitor is half filled with a dielectric with a dielectric constant of 2. The boundary between the dielectric and air is perpendicular to the plates. The distance between the plates is 1 cm, and the area of ​​the plates is 50 cm^2.

To solve the problem, it is necessary to determine the change in the energy of the electric field of the capacitor. To do this, you can use the formula for the energy of the electric field of a capacitor:

W = (1/2)CV^2,

where W is energy, C is capacitance of the capacitor, V is voltage across the capacitor.

The capacitance of the capacitor is determined by the formula:

C = εS/d,

where ε is the dielectric constant, S is the area of ​​the plates, d is the distance between the plates.

The first step is to determine the capacitance of the capacitor. Since the capacitor is half filled with dielectric, the dielectric constant must be taken into account when calculating the capacitance. Thus, the capacitance of the capacitor is:

C = εS/(2d) = (228.8510^-125010^-4)/(2110^-2) = 8.8510^-9 F.

The voltage across the capacitor can then be determined. Since the source is connected directly to the capacitor, the voltage across the capacitor will be equal to the emf. source, that is 2 V.

Now we can use the formula for the energy of the electric field of the capacitor:

W = (1/2)CV^2 = (1/2)8.8510^-9*(2)^2 = 8.85*10^-9 J.

Thus, the change in the energy of the electric field of a capacitor when it is half filled with a dielectric with a dielectric constant of 2, provided that the plates are connected to a source with an emf. 2 V equals 8.85*10^-9 J.

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