Physics, 29.02.2020 01:14 beej543

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

A) Find , the energy dissipated in the resistor.
Express your answer in terms of 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. (Part B figure) The battery is then disconnected and the capacitor discharged. For this situation, what is , the energy dissipated in the resistor?
Express your answer in terms of and other given quantities.

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Answer from: brenyasanders4001

The energy dissipated in the resistor ${U_k} = \frac{U}{k}$

The energy dissipated in the resistor ${U_k} = kU$

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{q ^2}{C}$

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation q = CV is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

$U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2} kCV ^2$

In this equation, is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2$

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Substitute $U = \frac{1}{2}\frac{{{q^2}}}{C}$ in the equation of ${U_k}$

$U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}$

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

$U_{k} = \frac{1}{2}kCV^2$

Here, V is voltage in the circuit.

Substitute $U =\frac{1}{2} CV^2$ in the equation of ${U_k}$

So,

$U_{k} = \frac{1}{2} kCV^2\\$

$= k(\frac{1}{2} CV^2)$

$U_{k} = kU$

Answer from: Quest

$p_{i}$ = initial pressure

$p_{f}$ = final pressure = 2 $p_{i}$

$v_{i}$ = initial volume

$v_{f}$ = final volume = $p_{i}$ /4

$t_{i}$ = initial temperature = 20 c = 20 + 273 = 293 k

$t_{f}$ = final temperature = ?

using the equation

$\frac{p_{i}v_{i}}{t_{i}}= \frac{p_{f}v_{f}}{t_{f}}$

$\frac{p_{i}v_{i}}{293} = \frac{(2p_{i}) (v_{i}/4)}{t_{f}}$

$\frac{1}{293} = \frac{0.5}{t_{f}}$

$t_{f}=10$

Answer from: Quest

The answer which is not a detail carson used to describe the pool in the cave at low tide is b. hundreds of periwinkles browse the branches and roots of the trees.

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