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
sweet [91]
3 years ago
5

A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie

ce of material having a dielectric constant of 3.55 is placed between the plates, completely filling the space between them.Part A. How much energy is stored in the capacitor before the dielectric is inserted?
Part B. How much energy is stored in the capacitor after the dielectric is inserted?
Part C. By how much did the energy change during the insertion?
Physics
1 answer:
-BARSIC- [3]3 years ago
8 0

a) 3.27\cdot 10^{-3} J

b) 11.60\cdot 10^{-3} J

c) 8.33\cdot 10^{-3} J

Explanation:

a)

The energy stored in a capacitor is given by

U=\frac{1}{2}CV^2

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

C=13.5 \mu F = 13.5\cdot 10^{-6}F is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J

b)

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

C'=kC

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2

where:

C=13.5\cdot 10^{-6}F is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J

C)

The initial energy stored in the capacitor, before the dielectric is inserted, is

U=3.27\cdot 10^{-3} J

The final energy stored in the capacitor, after the dielectric is inserted, is

U'=11.60\cdot 10^{-3} J

Therefore, the change in energy of the capacitor during the insertion is:

\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J

So, the energy of the capacitor has increased by 8.33\cdot 10^{-3} J

You might be interested in
Find acceleration. Will give brainliest!
zalisa [80]

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

7 0
3 years ago
Read 2 more answers
What is the strength of the electric field ep 0.90 mm from a proton?
dem82 [27]

The electric field strength is inversely related to the square of the distance.so  the strength of the electric field is given by

E=\frac{1}{4\pi \epsilon _{0}  } \frac{q}{r^{2} } = \frac{k q}{r^{2} }

Here,  \frac{1}{4\pi \epsilon _{0}  } = k  is constant depend upon medium and its value is 9.0 \times10^{9} \ N m^2/C^2 and q is charge and  r is the distance.

Given  r = 0.90 mm = 9.0 \times 10^{-4} m and we know the charge of proton, q = 1.6\times 10^{-19} \ C.

Therefore,

E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C  }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\  E= 1.77 \times 10^{-3}  N/C

4 0
3 years ago
A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationary seal of m
anzhelika [568]

Answer:

Momentum after collision will be 6000 kgm/sec

Explanation:

We have given mass of the whale = 1000

Initial velocity v = 6 m/sec

It collides with other mass of 200 kg which is at stationary

Initial momentum of the whale = 1000×6 = 6000 kgm/sec

We have to find the momentum after collision

From conservation of momentum

Initial momentum = final momentum

So final momentum = 6000 kgm/sec

5 0
3 years ago
Where would a car traveling on a roller coaster have the most kinetic energy ? and why?
goldenfox [79]

Answer:

As the car travels up the coaster it is gaining potential energy.

Explanation:

Because It has the greatest in amount of potential energy at the top of the coaster. when the car travels down the roller coaster it obtains speed and kinetic energy.

4 0
2 years ago
Read 2 more answers
State the laws of vibration of a stringed Instrument​
olchik [2.2K]
If the length and linear density are constant, the frequency is directly proportional to the square root of the tension.
7 0
2 years ago
Other questions:
  • Hey guys I really really need help with this question for ASAP! Explain what chart junk is and how it differs from the kind of i
    13·2 answers
  • Luke pulls a sled using a force of 65 N for a distance of 25 m across level ground. The rope Luke pulls forms an angle of 25° wi
    12·2 answers
  • What is the voltage in a circuit that has a current of 10.0 amps and a resistance of 28.5 ohms?
    5·2 answers
  • A force of 100 N is applied to a crate at an angle of 60° to the horizontal surface. The force causes the crate to move 2 m acro
    13·2 answers
  • How do transformers work in electric power houses
    14·1 answer
  • an object of mass 4 kg is placed on an inclined plane at an angle of 60. if the object slides down the plane with an acceleratio
    15·1 answer
  • A lightning bolt occurs when billions of protons are transferred at the same time. ____________________
    13·1 answer
  • Why do comets have tails?
    8·1 answer
  • What is the IMA of the following pulley system?
    9·2 answers
  • Lab: Waves and Diffraction<br> Assignment: Lab Report<br> Anyone have this completed
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!