The first charge is pulling on the second. Is, №17886436, 04.06.2020 00:12

Physics

The first charge is pulling on the second. Is the second pulling on the first? Explain your reasoning

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P Answered by PhD

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Physics

Aparallel-plate capacitor initially has air (k= 1) between the plates. we first fullycharge it by a 12 v battery. after the battery is disconnected, we insert a dielectricbetween the plates and it completely fills the space in between. a voltmeter is placedacross the capacitor and it reads 3.6 v now.(a) assuming that there is no charge loss during the process, what is the dielectricconstant of this material? (b) what’s the fraction of the stored energy changed by inserting the dielectric? (c) what will the voltmeter read if the dielectric is pulled

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P Answered by PhD

(a) 3.33

For a capacitor with dielectric disconnected from the battery, the relationship between the voltage across the capacitor without the dielectric (V) and with the dielectric (V) is given by

$V' = \frac{V}{k}$

where

k is the dielectric constant of the material

In this problem, we have

V' = 3.6 V

V=12 V

So we can re-arrange the formula to find the dielectric constant:

$k=\frac{V}{V'}=\frac{12 V}{3.6 V}=3.33$

(b) The energy stored reduces by a factor 3.33

The energy stored in a capacitor is

$U=\frac{1}{2}QV$

where

Q is the charge stored on the capacitor

V is the voltage across the capacitor

Here we can write the initial energy stored in the capacitor (without dielectric) as

$U=\frac{1}{2}QV$

while after inserting the dielectric is

$U'=\frac{1}{2}QV' = \frac{1}{2}Q\frac{V}{k}$

since Q, the charge, has not changed (the capacitor is disconnected, so the charge cannot flow away from the capacitor).

So the ratio between the two energies is

$\frac{U'}{U}=\frac{\frac{1}{2}Q\frac{V}{k}}{\frac{1}{2}QV}=\frac{1}{k}$

which means

$U' = \frac{U}{k}=\frac{U}{3.33}$

So, the energy stored has decreased by a factor 3.33.

(c) 5.5 V

Pulling the dielectric only partway so that it fills half of the space between the plates is equivalent to a system of 2 capacitors in parallel, each of them with area A/2 (where A is the original area of the plates of the capacitor), of which one of the two is filled with dielectric while the other one is not.

Calling

$C=\frac{\epsilon_0 A}{d}$ the initial capacitance of the capacitor without dielectric

The capacitance of the part of the capacitor of area A/2 without dielectric is

$C_1 = \frac{\epsilon_0 \frac{A}{2}}{d}= \frac{C}{2}$

while the capacitance of the part of the capacitor with dielectric is

$C_2 = \frac{k \epsilon_0 \frac{A}{2}}{d}= \frac{kC}{2}$

The two are in parallel, so their total capacitance is

$C' = C_1 + C_2 = \frac{C}{2}+\frac{kC}{2}=(1+k)\frac{C}{2}=(1+3.33)\frac{C}{2}=2.17 C$

We also have that

$V=\frac{Q}{C}=12 V$ this is the initial voltage

So the final voltage will be

$V' = \frac{Q}{C'}=\frac{Q}{2.17 C}=\frac{1}{2.17}V=\frac{12 V}{2.17}=5.5 V$

Physics

Aparallel-plate capacitor initially has air (k= 1) between the plates. we first fullycharge it by a 12 v battery. after the battery is disconnected, we insert a dielectricbetween the plates and it completely fills the space in between. a voltmeter is placedacross the capacitor and it reads 3.6 v now.(a) assuming that there is no charge loss during the process, what is the dielectricconstant of this material? (b) what’s the fraction of the stored energy changed by inserting the dielectric? (c) what will the voltmeter read if the dielectric is pulled

Step-by-step answer

P Answered by PhD

(a) 3.33

For a capacitor with dielectric disconnected from the battery, the relationship between the voltage across the capacitor without the dielectric (V) and with the dielectric (V) is given by

$V' = \frac{V}{k}$

where

k is the dielectric constant of the material

In this problem, we have

V' = 3.6 V

V=12 V

So we can re-arrange the formula to find the dielectric constant:

$k=\frac{V}{V'}=\frac{12 V}{3.6 V}=3.33$

(b) The energy stored reduces by a factor 3.33

The energy stored in a capacitor is

$U=\frac{1}{2}QV$

where

Q is the charge stored on the capacitor

V is the voltage across the capacitor

Here we can write the initial energy stored in the capacitor (without dielectric) as

$U=\frac{1}{2}QV$

while after inserting the dielectric is

$U'=\frac{1}{2}QV' = \frac{1}{2}Q\frac{V}{k}$

since Q, the charge, has not changed (the capacitor is disconnected, so the charge cannot flow away from the capacitor).

So the ratio between the two energies is

$\frac{U'}{U}=\frac{\frac{1}{2}Q\frac{V}{k}}{\frac{1}{2}QV}=\frac{1}{k}$

which means

$U' = \frac{U}{k}=\frac{U}{3.33}$

So, the energy stored has decreased by a factor 3.33.

(c) 5.5 V

Pulling the dielectric only partway so that it fills half of the space between the plates is equivalent to a system of 2 capacitors in parallel, each of them with area A/2 (where A is the original area of the plates of the capacitor), of which one of the two is filled with dielectric while the other one is not.

Calling

$C=\frac{\epsilon_0 A}{d}$ the initial capacitance of the capacitor without dielectric

The capacitance of the part of the capacitor of area A/2 without dielectric is

$C_1 = \frac{\epsilon_0 \frac{A}{2}}{d}= \frac{C}{2}$

while the capacitance of the part of the capacitor with dielectric is

$C_2 = \frac{k \epsilon_0 \frac{A}{2}}{d}= \frac{kC}{2}$

The two are in parallel, so their total capacitance is

$C' = C_1 + C_2 = \frac{C}{2}+\frac{kC}{2}=(1+k)\frac{C}{2}=(1+3.33)\frac{C}{2}=2.17 C$

We also have that

$V=\frac{Q}{C}=12 V$ this is the initial voltage

So the final voltage will be

$V' = \frac{Q}{C'}=\frac{Q}{2.17 C}=\frac{1}{2.17}V=\frac{12 V}{2.17}=5.5 V$

Physics

Mr. Mullenmeister is photocopying lab sheets for his first-period class when the toner light on the onA particle of toner carrying a charge of the copying machine experiences an electric field of 1.2 * 10 ^ 6 * N / C as it s pulled toward the paper. What is the electric force acting on the toner particle? F=4.5=(40*10^ C)(1.2*106N/C)=4.8*10^ .5

Step-by-step answer

P Answered by PhD

0.0048 N

Explanation:

The electric force acting on a particle is given by

F=qE

where

q is the charge

E is the magnitude of the electric force

F is the strength of the force

The force has the same direction as the field if the charge is positive, and the opposite direction if the charge is negative.

In this problem, we have:

$Q=4.0\cdot 10^{-9}C$ is the charge

$E=1.2\cdot 10^6 N/C$ is the electric field

Therefore, the electric force on the charge is

$F=QE=(4.0\cdot 10^{-9})(1.2\cdot 10^6)=0.0048 N$

Physics

Mr. Mullenmeister is photocopying lab sheets for his first-period class when the toner light on the onA particle of toner carrying a charge of the copying machine experiences an electric field of 1.2 * 10 ^ 6 * N / C as it s pulled toward the paper. What is the electric force acting on the toner particle? F=4.5=(40*10^ C)(1.2*106N/C)=4.8*10^ .5

Step-by-step answer

P Answered by PhD

0.0048 N

Explanation:

The electric force acting on a particle is given by

F=qE

where

q is the charge

E is the magnitude of the electric force

F is the strength of the force

The force has the same direction as the field if the charge is positive, and the opposite direction if the charge is negative.

In this problem, we have:

$Q=4.0\cdot 10^{-9}C$ is the charge

$E=1.2\cdot 10^6 N/C$ is the electric field

Therefore, the electric force on the charge is

$F=QE=(4.0\cdot 10^{-9})(1.2\cdot 10^6)=0.0048 N$

Physics

Question 1 According to the Law of Conservation of Energy, why does the first hill on a roller coaster always have to be the tallest of all the other hills? Question 2 A single marble is placed at the top of a hill. The initial potential energy of the marble is 550 J. The marble slides down the hill. You take a picture of the marble at some point as it is down the hill. Without knowing the speed, the mass or the position of the marble, which of the following could be the only possible correct energies of the marble at this point? Assume no friction

Step-by-step answer

P Answered by Master

4

Dang that’s a lot wait i will answer in comments

Mathematics

I will give brainleist to the first person who answers Mr. Smith needed to fill his truck with diesel, so he drove right to the nearest station and pulled in slowly. The gas panel was on the left side of his truck so he pulled up to an open pump. He put 20 gallons of diesel into the truck for a total charge of $48.60. When he went to pay, he put a $50 bill down on the counter. The clerk looked up and said “Here’s your change, have a great day.” “You too,” he said as he left to return to his truck. 6 Letter Direction Lock (use: U for UP, D for Down,

Step-by-step answer

P Answered by PhD

what is the question

the change is 1.40

Step-by-step explanation:

Physics

Question 1 According to the Law of Conservation of Energy, why does the first hill on a roller coaster always have to be the tallest of all the other hills? Question 2 A single marble is placed at the top of a hill. The initial potential energy of the marble is 550 J. The marble slides down the hill. You take a picture of the marble at some point as it is down the hill. Without knowing the speed, the mass or the position of the marble, which of the following could be the only possible correct energies of the marble at this point? Assume no friction

Step-by-step answer

P Answered by Master

4

Dang that’s a lot wait i will answer in comments

Mathematics

I will give brainleist to the first person who answers Mr. Smith needed to fill his truck with diesel, so he drove right to the nearest station and pulled in slowly. The gas panel was on the left side of his truck so he pulled up to an open pump. He put 20 gallons of diesel into the truck for a total charge of $48.60. When he went to pay, he put a $50 bill down on the counter. The clerk looked up and said “Here’s your change, have a great day.” “You too,” he said as he left to return to his truck. 6 Letter Direction Lock (use: U for UP, D for Down,

Step-by-step answer

P Answered by PhD

what is the question

the change is 1.40

Step-by-step explanation:

Physics

Mr. Patel is photocopying lab sheets for his first period class. A particle of toner carrying a charge of 4.0 * 10^9 C in the copying machine experiences an electric field of 1.2 * 10^6 N/C as it’s pulled toward the paper. What is the electric force acting on the toner particle?

Step-by-step answer

P Answered by PhD

2

4.8 × 10^15 N

Explanation:

Electric Field is defined as Force per unit Charge.

This is expressed mathematically as;

E= F/Q

Where E- Electric Field

F- Force

Q- charge

From the expression above by change of subject of formula for F, we have;

F=E×Q

= 1.2 * 10^6 ×4.0 * 10^9

= 4.8 × 10^15 N

The first charge is pulling on the second. Is the second pulling on the first? Explain your reasoning

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