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A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is U. The battery is removed, and then a dielectric material with dielectric constant K is inserted into the capacitor, filling the space between the plates. Finally, the capacitor is fully discharged through a resistor (which is connected across the capacitor terminals).

A.)Find U_{r}, the the energy dissipated in the resistor.

Express your answer in terms of U and other given quantities.

B.) Consider the same situation as in the previous part, except that the charging battery remains connected while the dielectric is inserted. The battery is then disconnected and the capacitor is discharged. For this situation, what is U_{r}, the energy dissipated in the resistor?

Express your answer in terms of U and other given quantities.

A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is U. The battery is removed, and then a dielectric material with dielectric constant K is inserted into the capacitor, filling the space between the plates. Finally, the capacitor is fully discharged through a resistor (which is connected across the capacitor terminals).

A.)Find U_{r}, the energy dissipated in the resistor. Express your answer in terms of U and other given quantities.

B.) Consider the same situation as in the previous part, except that the charging battery remains connected while the dielectric is inserted. The battery is then disconnected and the capacitor is discharged. For this situation, what is U_{r}, the energy dissipated in the resistor? Express your answer in terms of U and other given quantities.

Concepts and Reason

To resolve the issue that energy storage in the Parallel Plate capacitor can be the main idea.

First, select the first instance. After that, utilize the formula to determine the amount of energy that is stored in the parallel plate capacitor. Find the energy stored within the capacitor that is parallel to the plate without a dielectric. Then, determine the amount of energy stored inside the capacitor in the event that a dielectric is present. The equation will be transformed into the energy stored in the capacitor without the necessity of dielectric. In this case the equation must not be written in the form of voltage.

Part B employs the equation to calculate the energy lost by the resistor. The equation must be written in the format of an equation for the energy stored in the capacitor in the parallel plate. The equation should be written in the format of voltage, which tells you what the length of time that the battery will remain connected.

Fundamentals

The dielectric as well as the parallel plate capacitors are the storage of energy in the capacitor.

In the case of capacitors, their energy is equivalent to the energy that is lost to the resistor. This equation indicates that Uk is the energy that is lost in resistors. Q is the charge, and k is the dielectric constant. C is the capacitance.

The equation for energy storage in the capacitor with a parallel-plate without dielectric is no longer valid.

In this equation, in this equation, the term “U” is the term used to describe that energy contained in the capacitor that is parallel to the plate without dielectric. The term Q is charge , and C is capacitance.

Part B has the same configuration as part A. However, the battery remains connected. The equation Q = CVq = CV is, therefore, completed.

The energy stored by capacitors made of dielectric in parallel plates is therefore absorbed by

This equation illustrates that Uk is the energy that is lost to resistors. V is the voltage, and k is the dielectric constant. C is the capacitance.

The equation that describes the energy stored by the capacitor with a parallel-plate design without dielectric is

In this equation, in this equation is the energy that is dissipated by the resistance, the term V is the voltage, k the dielectric constant, and the C is the capacity.

To solve the problem, energy stored in a parallel plate capacitor is the key concept.

First, choose the first case. Then, use the formula to calculate the energy stored in a parallel plate capacitor. Find the energy stored in the parallel-plate capacitor without a dielectric. Next, calculate the energy stored within the capacitor when the dielectric is present. This equation will be converted to the energy stored inside the capacitor without the use of a dielectric. In this instance, the equation should not be written in the form voltage.

Part B uses the equation for the energy dissipated by the resistor. It should be written in the form of an equation for energy stored within the parallel plate capacitor. The equation should be written in the form voltage, which is how long the battery is connected.

The dielectric and the parallel plate capacitor are the energy stores in the capacitor.

[katex]{U_k} = \frac{1}{2}\frac{{{q^2}}}{{kC}}[/katex]

The energy stored in the capacitor is equal to the energy lost in the resistor. This equation shows that Uk is energy lost in resistors, q is charge and k is dielectric constant. C is the capacitance.

The equation of energy stored in the parallel-plate capacitor without dielectric is now.

[katex]U = \frac{1}{2}\frac{{{q^2}}}{C}[/katex]

In this equation

Urefers to the energy stored in the parallel-plate capacitor without dielectric.Qis charge andCis capacitance.Part B is the same as part A, but the battery is still connected. The equation [katex]q = CV[/katex] is thus completed.

The energy stored in parallel plate capacitors with dielectric is thus given by

[katex]\begin{array}{c}\\{U_k} = \frac{1}{2}\frac{{{k^2}{C^2}{V^2}}}{{kC}}\\\\ = \frac{1}{2}kC{V^2}\\\end{array}[/katex]

This equation shows that Uk is energy lost in resistors, V is voltage and k is dielectric constant. C is the capacitance.

The equation for the energy stored in the parallel-plate capacitor without dielectric is

[katex]\begin{array}{c}\\U = \frac{1}{2}\frac{{{C^2}{V^2}}}{C}\\\\ = \frac{1}{2}C{V^2}\\\end{array}[/katex]

In this equation

Urefers to the energy dissipated by the resistor,Vis voltage,kthe dielectric constant andCthe capacitance.(A)The energy dissipated by the resistor is described in this equation:

[katex]{U_k} = \frac{1}{2}\frac{{{q^2}}}{{kC}}[/katex]

Substitute [katex]U = \frac{1}{2}\frac{{{q^2}}}{C}[/katex]

[katex]\begin{array}{c}\\{U_k} = \frac{1}{2}\left( {\frac{1}{k}} \right)\frac{{{q^2}}}{C}\\\\ = \left( {\frac{1}{k}} \right)\frac{{{q^2}}}{{2C}}\\\\{U_k} = \frac{U}{k}\\\end{array}[/katex]

[Part A]

Part APart A(B)The energy dissipated by the resistor is described in this equation:

[katex]{U_k} = \frac{1}{2}kC{V^2}[/katex]

Vrefers to the voltage within the circuit.Substitute [katex]U = \frac{1}{2}C{V^2}[/katex]

[katex]U = \frac{1}{2}\frac{{{q^2}}}{C}[/katex]0

[Part B]

Part APart AAns Part A

The resistor [katex]U = \frac{1}{2}\frac{{{q^2}}}{C}[/katex]1Part A dissipates the energy

The resistor [katex]U = \frac{1}{2}\frac{{{q^2}}}{C}[/katex]2 dissipates the energy