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
WINSTONCH [101]
3 years ago
15

A 13.0 μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 VV across the plates. A pi

ece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them.
(a) How much energy is stored in the capacitor before and after the dielectric is inserted?
(b) By how much did the energy change during the insertion? Did it increase or decrease?

Physics
2 answers:
horsena [70]3 years ago
8 0

Answer:

Energy stored in the capacitor before the dielectric was inserted=

3.744×10-3 J

Energy stored in the capacitor after the dielectric was inserted=

14.0×10-3 J

Change in energy due to dielectric

10.3×10-3 J or 10mJ

The energy stored in the capacitor increased due to the insertion of the dielectric material.

Explanation:

The energy stored in the capacitor is calculated from the formula

U= 1/2CV^2

Where:

U=energy stored in the capacitor

C=capacitance of the capacitor

V= voltage.

The energy stored in the capacitor before the insertion of the dielectric is shown as U while the energy stored in the capacitor after the insertion of the dielectric is designated U'.

Since U'>U from the solution attached, the change in energy due to insertion of the dielectric ∆U= U'-U

Hence the answer shown. See attached image for detailed solution.

mestny [16]3 years ago
7 0

Given Information:

Capacitance = C = 13  μF

Potential difference = V = 24 V

Dielectric constant = de = 3.75

Required Information:

(a) Energy stored before and after adding dielectric = ?

(b) change in energy = ?

Answer:

(a) E₁ = 3.74x10⁻³ F and E₂ = 14.04x10⁻³ F

(b) ΔE = 10.3x10⁻³ F increased

Explanation:

The energy stored in a capacitor is given by

E = ½CV²

Where C is the capacitance and V is the potential difference

Energy before the dielectric:

E₁ = ½CV²

E₁ = ½(13x10⁻⁶)(24)²

E₁ = 3.74x10⁻³ F

Energy after the dielectric:

E₂ = ½CdeV²

E₂ = ½(13x10⁻⁶)(3.75)(24)²

E₂ = 14.04x10⁻³ F

Change in energy:

ΔE = E₂ - E₁

ΔE = 14.04x10⁻³ - 3.74x10⁻³

ΔE = 10.3x10⁻³ F

The energy increased after adding the dielectric material.

You might be interested in
****PLEASE HELP**** THERE ARE TWO QUESTIONS (ITS EASY)
notka56 [123]

but I think it's a and f

and

b and e

3 0
3 years ago
QUESTION 1
Molodets [167]
Question 1: C Question 2: B, Hope this Helps!
3 0
3 years ago
a block that is 80 degrees is places next to a 20 degree block. the warmer block heats the cooler block by the process of... A)c
11Alexandr11 [23.1K]
Answer: option B: conduction.

Conduction is the heat transfer that happens between two bodies in direct contact, due to the collision of the molecules, atoms and electrons within the body (microscopical level).
6 0
3 years ago
How are electromagnetic waves able to travel without a medium?
vazorg [7]
Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. this means that electromagnetic waves can travel not only trough air and solid materials, but also trough the vacuum of space.
7 0
3 years ago
Neutron stars consist only of neutrons and have unbelievably high densities. a typical mass and radius for a neutron star might
Tanya [424]
<span>Density is 3.4x10^18 kg/m^3 Dime weighs 1.5x10^12 pounds The definition of density is simply mass per volume. So let's divide the mass of the neutron star by its volume. First, we need to determine the volume. Assuming the neutron star is a sphere, the volume will be 4/3 pi r^3, so 4/3 pi 1.9x10^3 = 4/3 pi 6.859x10^3 m^3 = 2.873x10^10 m^3 Now divide the mass by the volume 9.9x10^28 kg / 2.873x10^10 m^3 = 3.44588x10^18 kg/m^3 Since we only have 2 significant digits in our data, round to 2 significant digits, giving 3.4x10^18 kg/m^3 Now to figure out how much the dime weighs, just multiply by the volume of the dime. 3.4x10^18 kg/m^3 * 2.0x10^-7 m^3 = 6.8x10^11 kg And to convert from kg to lbs, multiply by 2.20462, so 6.8x10^11 kg * 2.20462 lb/kg = 1.5x10^12 lb</span>
4 0
3 years ago
Other questions:
  • -<br> Speed is a scalar, a quantity that is<br> described by<br> alone.
    8·1 answer
  • According to a relatively recent discovery which member of the solar system has one satellite
    7·2 answers
  • Which of the following is not true about a good streching routines
    9·1 answer
  • There is a girl pushing on a large stone sphere. The sphere has a mass of 8200 kgand a radius of 90 cm and floats with nearly ze
    15·1 answer
  • What is the largest-aperture Earth-based telescope currently in use at visible wavelengths?
    15·1 answer
  • The greater the amplitude, the greater the wave's:
    9·1 answer
  • Vector e is 0.111m long in a 90 deg direction. Vector f is 0.233 m long in 400 deg direction. Find the direction of their vector
    13·1 answer
  • Please Help Me!!!
    13·2 answers
  • Hello guys I have a quick question and please help me! Thanks!
    11·2 answers
  • Am 1 years old UwU (JK 15)
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!