1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ELEN [110]
3 years ago
12

A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem

oved, 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.
Physics
1 answer:
Viktor [21]3 years ago
4 0

Answer:

A

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

B

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, Uk 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

You might be interested in
At Alpha Centauri's surface, the gravitational force between Alpha Centauri and a 2 kg mass of hot gas has a magnitude of 618.2
sergeinik [125]

Answer:

6.86 * 10^8 m

Explanation:

Parameters given:

Mass of hot gas, m = 2 kg

Gravitational Force, F = 618.2 N

Mass of Alpha Centauri, M = 2.178 * 10^30 kg

The gravitational force between two masses (the hot gas and Alpha Centauri) , m and M, at a distance, r, given as:

F = (G*M*m) / r²

Where G = gravitational constant

Therefore,

618.2 = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / r²

=> r² = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / 618.2

r² = 4.699 * 10^17 m²

=> r = 6.86 * 10^8 m

We are told that the hot gas is on the surface of Alpha Centauri, hence, the distance between both their centers is the radius of Alpha Centauri.

The mean radius of Alpha Centauri is 6.86 * 10^8 m.

5 0
3 years ago
Read 2 more answers
Two things you can do to increase the acceleration of an object
stiv31 [10]
You can decrease the mass, or you can increase the force applied to the object
7 0
3 years ago
The parietal lobe is the portion of the cerebral cortex located below the temporal lobe.
jasenka [17]

Answer:

False

Explanation:

Please see the attached file

5 0
3 years ago
Read 2 more answers
A proton with a speed of 3.2 x 106 m/s is shot into a region between two plates that are separated by a distance of 0.23 m. As t
harina [27]

Answer:

<h2>The magnitude of the magnetic is 0.145 T</h2>

Explanation:

Given :

Speed of proton v = 3.2 \times 10^{6}\frac{m}{s}

Mass of proton m = 1.67 \times 10^{-27} Kg

The force on the proton in magnetic field is given by,

  F = q (v \times B)

  F = qvB \sin \theta

But \sin 90 = 1    (∵ Force is perpendicular to the velocity so \theta = 90)

  F = qvB

When particle enter in magnetic field at the angle of 90° so particle moves in circle

So force is given by,

  F = \frac{mv^{2} }{r}

Where r = radius but in our case 0.23 m, q = 1.6 \times 10^{-19} C

By comparing above two equation,

  B = \frac{mv}{qr}

  B = \frac{1.67 \times 10^{-27} \times 3.2 \times 10^{6}  }{1.6 \times 10^{-19} \times 0.23 }

  B = 0.145 T

5 0
3 years ago
What is the speed of a bobsled whose distance-time graph indicates that it traveled 122m in 27s?
Stolb23 [73]
Speed = distance/time
speed= 122÷27=4.52m/s (3sf)
7 0
3 years ago
Other questions:
  • Astronomers estimate that a galaxy contains a. 100 stars. c. millions or billions of stars. b. 1 star. d. 100,000 stars.
    14·1 answer
  • On which factors does friction depend ?​
    5·1 answer
  • What is the first step for both loading and unloading a firearm?
    11·1 answer
  • Coin-shaped compartment that contains light-absorbing molecules
    6·1 answer
  • A man drops a baseball from the top of a building. If the ball is held at a height of 1m before it is dropped, and takes 6.8 sec
    11·1 answer
  • (a) Find the speed of waves on a violin string of mass 717 mg and length 24.3 cm if the fundamental frequency is 980 Hz. (b) Wha
    11·1 answer
  • An elevator filled with passengers has a mass of 1663 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 f
    9·1 answer
  • You have two small spheres, each with a mass of 2.40 grams, separated by a distance of 10.0 cm. You remove the same number of el
    13·1 answer
  • 3. How do you record your score in the 3-Minute Step Test? A record the 60-second heart rate before activity B. record the 30-se
    8·1 answer
  • What direction would the north pole of a bar magnet point if you were to hang the bar magnet from a thin string?.
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!