Physics question thread

[Shrike]

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All users may post questions about MCAT and OAT physics here. We will answer the questions as soon as we reasonably can. If you would like to know what physics topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm), though be warned, there are subjects listed there that are rarely tested, or that appear in passages only and need not be learned.

Be sure to check the Physics FAQs and Topic Writeups thread if you have a general question; eventually, many of your answers will be located there. Also, a request: to keep this thread at least somewhat neat, when replying to someone else's post please refrain from quoting anything more than what's necessary for clarity.

Acceptable topics:

• general, MCAT-level physics
• particular MCAT-level physics problems, whether your own or from study material
• what you need to know about physics for the MCAT
• how best to approach to MCAT physics passages
• how best to study MCAT physics
• how best to tackle the MCAT physical sciences section

Unacceptable topics:

• actual MCAT questions or passages, or close paraphrasings thereof
• anything you know to be beyond the scope of the MCAT

Side note: anyone who knows how to post subscripts and superscripts in this system, please PM me the method. I also wouldn't mind knowing how to post some obvious symbols, such as Greek letters and an infinty sign. Should be a matter of changing fonts, but I don't know what's available; again, a PM would be appreciated.

If you really know your physics, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current official contributors to the this thread -- a team to which I hope to add more people:

Thread moderated by: Shrike. Shrike is a full-time instructor for The Princeton Review; he has taken the MCAT twice for no good reason, scoring 14 on the physical sciences section each time. He majored in mathematics, minored in physics, and spent several years accumulating unused school experience (in economics and law).

Also answering questions: Xanthines, a Kaplan instructor. He scored 13 on the PS section of the MCAT and 34 overall.

 

[87138]

Speaking of electrons and fields . . .

Ok this question is linked to a passage (not AAMC) about charged oil droplets in an electric field. The background really isn't necessary I don't think.

A drop of oil of mass 5x10^-16 kg is at rest on the bottom of a plate of a parallel plate combination when the electric field is zero. An electric field of 4x10^3 V/m is then applied between the plates, accelerating the drop towards the top plate. What will be the resultant acceleration of the drop if it carries a negative charge of 3x10^-18 C?

A) 9.8 m/s^2
B) 14.2 m/s^2
C) 24.0 m/s^2
D) 28.4 m/s^2


I answered C, but the "correct" answer is B. Here was my logic.

F = Eq, and F = ma, so Eq = ma.

(4x10^3)(3x10^-18) = (5x10^-16)(a)


Solve for a, and you get 24 on the dot.

Where am I going wrong here?

Anyone? 🙂

 

[gridiron]

Speaking of electrons and fields . . .

Ok this question is linked to a passage (not AAMC) about charged oil droplets in an electric field. The background really isn't necessary I don't think.

A drop of oil of mass 5x10^-16 kg is at rest on the bottom of a plate of a parallel plate combination when the electric field is zero. An electric field of 4x10^3 V/m is then applied between the plates, accelerating the drop towards the top plate. What will be the resultant acceleration of the drop if it carries a negative charge of 3x10^-18 C?

A) 9.8 m/s^2
B) 14.2 m/s^2
C) 24.0 m/s^2
D) 28.4 m/s^2


I answered C, but the "correct" answer is B. Here was my logic.

F = Eq, and F = ma, so Eq = ma.

(4x10^3)(3x10^-18) = (5x10^-16)(a)


Solve for a, and you get 24 on the dot.

Where am I going wrong here?

Grape was right!

 

[grapeflavorsoda]

Speaking of electrons and fields . . .

Ok this question is linked to a passage (not AAMC) about charged oil droplets in an electric field. The background really isn't necessary I don't think.

A drop of oil of mass 5x10^-16 kg is at rest on the bottom of a plate of a parallel plate combination when the electric field is zero. An electric field of 4x10^3 V/m is then applied between the plates, accelerating the drop towards the top plate. What will be the resultant acceleration of the drop if it carries a negative charge of 3x10^-18 C?

A) 9.8 m/s^2
B) 14.2 m/s^2
C) 24.0 m/s^2
D) 28.4 m/s^2


I answered C, but the "correct" answer is B. Here was my logic.

F = Eq, and F = ma, so Eq = ma.

(4x10^3)(3x10^-18) = (5x10^-16)(a)


Solve for a, and you get 24 on the dot.

Where am I going wrong here?

you forgot gravity.

if you define above ground as positivie direction, then

Eq - mg = ma

so a = (Eq - mg)/m

plug in and it should give the correct answer.

 

[gridiron]

If the electron is going right, --->, then the electric field would be pointing the opposite way? I'm just having trouble determining where the electric field points and my books say to use electric field lines but I don't know how.

Maybe a explanation will help 🙂 :
Place a sphere of negative charge in space. Now place a positive test charge near the sphere and you will see a electrostatic force will act on the test charge and point toward the center of the sphere. This means the electric field vectors, which is a property of the electric field given by the equation: E = Felectrostatic/charge, are directed radially toward the sphere. Field lines can be drawn to represent the vectors, but remember the relation between field lines and electric field vectors:

1.) At any point, the tangent to a curved field line or the direction of a straight field line is the direction of the electric field at that point.

2.) The field lines are drawn to represent the magnitude of the electric field at a particular point. Hence, when the field lines are close together, the electric field is large and when they are more spread apart, the electric field is small.

​

Now if the sphere was a positive charge, the electric field vectors will point radially away from the sphere.
What is important to remember is the electric field vectors and field lines are representations of what a positive test charge will feel in the presence of a like or dissimilar charge. I hope this helps and clears up any confusion!!

 

[87138]

Hmmm....interesting....from the question, I would have approached it the same way as you did. If it is a pair of parallel plates, than there is a constant electrostatic force acting upward on the charged drop---this leads to a constant acceleration in that direction. Very interesting......I am inclined to say the solution in the book is wrong, but perhaps you can elaborate on the solution given by the test?? Did the passage say the electric field is uniform and produced at all points between the plates? Did the passage mention any fringing of the electric field at the edges of the plates?

It was a Kaplan Physical Sciences ~30 minute test, and I only have the answer key, no solutions associated with it. For some reason I can't find it (I have a huge pile of MCAT-related papers in my room), but I don't recall any special conditions like fringing, and I remember thinking that all signs suggested a uniform electric field. I'm also inclined to believe the "correct" answer is a typo. It's at least nice to know that someone else can vouch for the approach I took towards the problem.

 

[gridiron]

you forgot gravity.

if you define above ground as positivie direction, then

Eq - mg = ma

so a = (Eq - mg)/m

plug in and it should give the correct answer.

I believe this is incorrect if there is a uniform electric field. Consider how a ink jet printer works. Drops are shot out of a generator and become charged in a charger. The computer controls the charge given to the drop. The drop then encounter a pair of conducting plates which are charged to produce a uniform electric field between them. Any charge will move upward at a constant acceleration. Usually, in cases such as these, the gravitational force on the drop is small relative to the electrostatic force and can be neglected---especially when the electric field is uniform.

 

[grapeflavorsoda]

I believe this is incorrect if there is a uniform electric field. Consider how a ink jet printer works. Drops are shot out of a generator and become charged in a charger. The computer controls the charge given to the drop. The drop then encounter a pair of conducting plates which are charged to produce a uniform electric field between them. Any charge will move upward at a constant acceleration.

it does move up at constant acceleration even if there is gravity(unless you fluctuate gravity somehow).

you normally ignore gravitational force when ratio of charge/mass is high.

but in this case charge/mass density is not that great, so you cannot ignore.

 

[grapeflavorsoda]

I believe this is incorrect if there is a uniform electric field. Consider how a ink jet printer works. Drops are shot out of a generator and become charged in a charger. The computer controls the charge given to the drop. The drop then encounter a pair of conducting plates which are charged to produce a uniform electric field between them. Any charge will move upward at a constant acceleration. Usually, in cases such as these, the gravitational force on the drop is small relative to the electrostatic force and can be neglected---especially when the electric field is uniform.

by the way, charge density of ink drop is normally constant in ink jet printer.

it just varies the electric field.

 

[87138]

you forgot gravity.

if you define above ground as positivie direction, then

Eq - mg = ma

so a = (Eq - mg)/m

plug in and it should give the correct answer.

Wow, can't believe I forgot gravity (or that I missed your post before I replied to BioMed's).

Thanks. That does the trick.

 

[gridiron]

by the way, charge density of ink drop is normally constant in ink jet printer.

it just varies the electric field.

An input signal from the computer controls the charge given to each drop and thus the effect of the electric field on the drop.

 

[87138]

Quick and dirty one: When you increase the distance between two plates of a parallel plate capacitor, does capicitance increase or decrease?

 

[deleted106503]

Quick and dirty one: When you increase the distance between two plates of a parallel plate capacitor, does capicitance increase or decrease?

Decreases because voltage goes up.

 

[deleted106503]

BioMed: For a/c circuits, does the MCAT test on any direct knowledge other than Vmax and Vrms relationship? EK didn't go over any of this except the sqrt of 2 x Vrms.

 

[gridiron]

BioMed: For a/c circuits, does the MCAT test on any direct knowledge other than Vmax and Vrms relationship? EK didn't go over any of this except the sqrt of 2 x Vrms.

Let me PM you the specifics!!!

 

[grapeflavorsoda]

An input signal from the computer controls the charge given to each drop and thus the effect of the electric field on the drop.

what i am saying is that it is typically the electrical field between deflecting electrodes is controlled not the charge density of the droplet.

the current burst model makes the ink droplet too small so it's production per sec must be quite large. and controlling ink charge density is more cumbersome and error-prone than varing deflecting electrodes' electric field.

hence when you actually measure the charge density of the ink droplets that are shoot out undeflected(no voltage difference between the deflecting electrodes), these undeflected ink droplet will have the same charge density as the droplets that are deflected.

 

[87138]

Let me PM you the specifics!!!

Feel free to send it my way too if you feel so inclined!

 

[gridiron]

Feel free to send it my way too if you feel so inclined!

Done!

 

[okcomputer]

I am a bit confused about something.

I understand that sound movement depends on the medium it goes through (tension, density, etc)

but what properties does light depend on besides temperature? and are there equations like the ones for sound(I know we wouldn't have to know them, but I usually learn better with visual aids)


thanks for any help

 

[lastbastion]

Ok, so Capacitance is equal to Charge divided by Voltage. A good capacitor, thus, is able to hold lots of gooey charge goodness at low voltage.

My question is this: Why then does capacitance not increase when resistance of the circuit it belongs to is reduced? I understand that capacitance depends on the area of the plates and the distance between the plates, from what I've read in subject books and test explanations. But, reducing resistance also seems like a plausible way to maximize capacitance. So, my line of thinking is faulty.

Who can show me where I am wrong?

 

[87138]

Done!

Thanks🙂.

 

[MedGrl@2022]

Somebody help me out... I know the equation... is there any good way to remember all the stuff about lights and optics... real vs. virtual... I get so confused... real is inverted... virtual is upright... real is positive... virtual is negative... is there some easy way to remember all of the stuff???

 

[doctor.nick]

can someone please post a little about how velocity, wavelength and frequency change as a wave is reflected

i remember that if a wave is reflected off an object moving towards the reciever, frequency increases

more specifically, i remember either velocity or wavelength do not change.. but i cant find my notes on that part

 

[Nutmeg]

can someone please post a little about how velocity, wavelength and frequency change as a wave is reflected

i remember that if a wave is reflected off an object moving towards the reciever, frequency increases

more specifically, i remember either velocity or wavelength do not change.. but i cant find my notes on that part

Velocity does not change. Velocity is only ever determined by the media/type of wave. A light wave will never change speed, and for sound or water, it has to do with the properties of the medium (density, viscosity, temperature, etc: things that won't change because of a reflecting surface). And with a constat velocity, a change in frequency gives a change in wavelength (higher frequency = shorter wavelength, lower frequency = longer wavelength).

 

[Foolins]

I'm not sure that's right....i thought that the velocity of a light wave can change speed to differing degrees based on the wavelength. That would explain why shorter wavelengths have a higher index of refraction upon entering a more dense medium

 

[doctor.nick]

thanks for the responses

my question was not regarding a change in media.. i believe foolins is thinking of snells law: c = nv

 

[Foolins]

No, i'm thinking of the properties of light that allow for rainbows. The violet end of the spectrum bends more (towards or away from the normal upon changing mediums).

Also, remember that frequency never changes, so when light slows down or speeds up, the wavelength must change a bit.

 

[deleted106503]

BioMed, thanks for all the help. I really appreciate it! 👍 All the work paid off today!

 

[QofQuimica]

BioMed, thanks for all the help. I really appreciate it! 👍 All the work paid off today!

lmao, I saw you posting in here, and I thought I was going to have to give you a talking-to! 😛 You're welcome, williams; hope you get the score you want, and good luck with your apps. 🙂

 

[deleted106503]

lmao, I saw you posting in here, and I thought I was going to have to give you a talking-to! 😛 You're welcome, williams; hope you get the score you want, and good luck with your apps. 🙂

I honestly have ZERO questions about my exam today. I pretty much forget it all already, lol! I'm already forgetting my EK material!

 

[gridiron]

I honestly have ZERO questions about my exam today. I pretty much forget it all already, lol! I'm already forgetting my EK material!

Congratulations WilliamsF1 🙂. I hope you did well today and good luck with your applications. I am sure you did well on this historic day----no more paper MCAT!!! Best of luck and feel free to ask me questions if you need any help.

P.S. Go BULLS!!!!!!!

 

[mcatcrazy]

Yes, it is finally over. Biomedengineer, thank you for all your physics help and for taking the time to pm with answers to my questions. Your help really paid off in the end. Thank you!!

 

[87138]

Yeah, thanks BioMed, I appreciate the help. Even though I don't feel that I did as well on PS as in my practice exams, but we'll see!

 

[gridiron]

Yes, it is finally over. Biomedengineer, thank you for all your physics help and for taking the time to pm with answers to my questions. Your help really paid off in the end. Thank you!!

You're Welcome 🙂 . I am glad that I was of help. Best of to you and I am sure you did well. Just be positive!! Good luck with med school applications!!

 

[gridiron]

Yeah, thanks BioMed, I appreciate the help. Even though I don't feel that I did as well on PS as in my practice exams, but we'll see!

You're welcome 🙂 (Even though I answered one of your questions incorrectly at first 🙁--apologies) I am sure you did well and believe in yourself. Stay positive and don't let the wait get to you--I know this is easier said then done. Your hard work and sacrifice will pay off. Best of luck!!!

 

[ih8mcat]

REPOST, BEYOND THE SCOPE OF THE MCAT: This is a relativity problem, I'm supposed to start by finding the velocity, but i'm not sure if I use the earth's circumference as the distance for the v=d/t...ok here's the problem:

A clock is placed in a satellite that orbits the earth with a period of 119 min. By what time interval (in seconds) will this clock differ from an identical clock on earth after 1 y? (Assume that special relativity applies and neglect general relativity.)

Any help would be greatly appreciated, thanks!

 

[hmcalley]

This is a question from NOVA... is this okay to post?

I don't get why D is wrong.

Reynold’s number is given by Re=lpv/n, where l is the length scale of the pipe or of obstacles, p is the density f the fluid, v is the velocity of the fluid, and n is the viscosity. The greater Reynold’s number for a given flow, the more likely it is that turbulence will develop. Consider water flowing in a pipe. Which of the following would tend to reduce the likelihood of turbulent flow?
A. Increase the flow rate.
B. Make the joints in the pipe smooth.
C. Raise the temperature
D. Increase the radius of the pipe.

I don't understand why is it that increasing the radius of a pipe for a given flow would tend to increase the likelihood of turbulent flow. Since flow rate= area x velocity, wouldn’t increasing the radius of the pipe increase the area and therefore decrease velocity ... so to achieve a lower Reynold’s number... so less likely for turbulence to develop?

Help? Thanks.

 

[gridiron]

This is a question from NOVA... is this okay to post?

I don't get why D is wrong.

Reynold’s number is given by Re=lpv/n, where l is the length scale of the pipe or of obstacles, p is the density f the fluid, v is the velocity of the fluid, and n is the viscosity. The greater Reynold’s number for a given flow, the more likely it is that turbulence will develop. Consider water flowing in a pipe. Which of the following would tend to reduce the likelihood of turbulent flow?
A. Increase the flow rate.
B. Make the joints in the pipe smooth.
C. Raise the temperature
D. Increase the radius of the pipe.

I don't understand why is it that increasing the radius of a pipe for a given flow would tend to increase the likelihood of turbulent flow. Since flow rate= area x velocity, wouldn’t increasing the radius of the pipe increase the area and therefore decrease velocity ... so to achieve a lower Reynold’s number... so less likely for turbulence to develop?

Help? Thanks.

The derivation of Reynolds number involves differentiating between inertial forces and viscous forces. Inertial forces are governed by the fluid velocity, characteristic length through which fluid flow occurs and the density of the fluid. Viscous forces are governed by kinematic fluid velocity, which is the ratio of the dynamic fluid viscosity to the density of the fluid. Overall, the Reynolds number is dimensionless and is the ratio of the inertial forces to the viscous forces. In most fluid dynamics problems, flow is produced due to a pressure gradient. In this case, it is important to understand what the characteristic length signifies in the problem. The characteristic length is given by the diameter if the cross section is circular. The diameter is given by D=2r, where r is the radius. So, if the radius of the pipe is increased, the characteristic length is increased. Looking at the ratio, this would increase the Reynolds number. Reynolds number can also be expressed in the form of the Darcy friction factor, usually applicable at low velocities in a cylindrical tube, as Re= (2*density*average velocity*radius/viscosity). An increase in the radius leads to an increase in the Reynolds number. I hope this helps.

 

[neurodoc]

This is a question from NOVA... is this okay to post?

I don't get why D is wrong.

Reynold’s number is given by Re=lpv/n, where l is the length scale of the pipe or of obstacles, p is the density f the fluid, v is the velocity of the fluid, and n is the viscosity. The greater Reynold’s number for a given flow, the more likely it is that turbulence will develop. Consider water flowing in a pipe. Which of the following would tend to reduce the likelihood of turbulent flow?
A. Increase the flow rate.
B. Make the joints in the pipe smooth.
C. Raise the temperature
D. Increase the radius of the pipe.

I don't understand why is it that increasing the radius of a pipe for a given flow would tend to increase the likelihood of turbulent flow. Since flow rate= area x velocity, wouldn’t increasing the radius of the pipe increase the area and therefore decrease velocity ... so to achieve a lower Reynold’s number... so less likely for turbulence to develop?

Help? Thanks.

I think B is the correct answer. The Reynold's equation does not have a factor that includes "radius" of pipe. It does have a factor, l, that includes length of pipe, including "obstacles" in the wall, such as joints. Turbulance is an edge effect upon laminar flow, and this effect results from the frictional interaction of the fluid with the walls of the pipe. Smoother walls will decrease l, resulting in a lower Reynold's number. Increasing the diameter of the pipe will increase its cross sectional area. It will also increase the surface area of the wall through which the fluid flows, thus increasing the friction and, to that extent, increasing the chance of turbulence at the wall of the pipe. In some real situations increasing the pipe diameter woould result in slower fluid flow (decreasing the chance of turbulence), but this would depend on factors not clear in the question.

Nick

 

[hmcalley]

Thanks biomedengineer and neurodoc! With your explanations, this problem is actually making sense : )

I have another question though. I just started the chapter on electrostatics. I read it twice now and still not really getting it... especially about voltage and work.

Here's the question. (can you tell me how you arrive at the answer? ... ie the equations to use? Why those equations?) Also, what does doing postive and negative work mean?

The negative charge (=-1μC) in the figure below goes from y=-5 to y=+5 and is made to follow the arrows in the vicinity of the two equal positive charges (=+5). What is the work required to move the negative charge along the arrows from y=-5 to +5? (For this part, the back of the book says no net work is done.) What is the work required to move the negative charge from y=-5 to y=0? (I don't see an answer to this one in the book..)

(Please see the linked figure.)

!! Don't get this physics : (

 

[Swenis]

I'm working on some chapter questions for my intro physics course (non-calc based), and I have managed to get every single one except for this one. Here's the question:

"Two masses are connected by a light string passing over a light, frictionless pulley, as shown above. Let

m1 = 8.0kg
m2 = 9.0kg
h = 40.0cm.

If both masses are released from rest, determine the speed of m1 just as m2 hits the ground"

I've done problems similar to this before, but I can't figure it out for the life of me. Any guidance would be awesome!

 

[gridiron]

I'm working on some chapter questions for my intro physics course (non-calc based), and I have managed to get every single one except for this one. Here's the question:

"Two masses are connected by a light string passing over a light, frictionless pulley, as shown above. Let

m1 = 8.0kg
m2 = 9.0kg
h = 40.0cm.

If both masses are released from rest, determine the speed of m1 just as m2 hits the ground"

I've done problems similar to this before, but I can't figure it out for the life of me. Any guidance would be awesome!

In order to solve problems of this type, it is crucial that you draw a free body diagram. For mass two, you have two forces acting on it: tension and gravity. The same goes for mass one. How do you determine direction? If you were to release the system from rest, which direction would it follow? Since mass 2 is greater than mass one, mass two would move down and mass one up. I determined this because the mass for each object was given in the problem. In the instance the mass was not given, you can arbitrarily choose the direction making note of signs and you should get the same result. For the two masses, use newton's second law, and you should get:

1.) T-m1g=m1a
2.) m2g-t=m2a

Why I choose the signs? I decided that moving down was positive for mass 2. For mass 1, gravity wants the mass to move down, but tension moves it up because of the direction of movement of mass 2. After some manipulation of this system of equations, you should obtain:

a= g(m2-m1)/m1+m2

This is the acceleration of the system. It should be positive in this case since mass 2 is greater than mass 1. The best way to check the answer is to understand this: if the masses were equal to one another, the system would be at rest. In the above equation, if we substitute m2=m1, we obtain an acceleration of zero, so our solution is correct. The system is at rest initially, so v initial is equal to zero. Therefore, you can use the following equation to solve for the final velocity:

(v final)^2 = (v initial)^2 + 2ad
Where d is the distance travelled. Mass 1 will travel a distance of h when mass 2 hits the ground. Substituting zero for v initial, the final velocity is equal to:

square root(2ah). That should be the answer. Make sure the units match. The height is given in centimeters, a cgs unit. Stick with SI units and convert the height to meters and you should obtain the correct answer.

I hope this helps. Good luck!

 

[hmcalley]

I posted this question a few days ago. Can someone help? I don't know how to post the figure, so I pasted it in a MS Word file.

The negative charge (=-1μC) in the figure below goes from y=-5 to y=+5 and is made to follow the arrows in the vicinity of the two equal positive charges (=+5). What is the work required to move the negative charge along the arrows from y=-5 to +5? (For this part, the back of the book says no net work is done.) What is the work required to move the negative charge from y=-5 to y=0? (I don't see an answer to this one in the book..)

(Please see the linked figure in post #589.)

Can you tell me how you arrive at the answer? ... ie the equations to use? Why those equations? Also, what does doing postive and negative work mean?


Thanks!

 

[gridiron]

I posted this question a few days ago. Can someone help? I don't know how to post the figure, so I pasted it in a MS Word file.

The negative charge (=-1μC) in the figure below goes from y=-5 to y=+5 and is made to follow the arrows in the vicinity of the two equal positive charges (=+5). What is the work required to move the negative charge along the arrows from y=-5 to +5? (For this part, the back of the book says no net work is done.) What is the work required to move the negative charge from y=-5 to y=0? (I don't see an answer to this one in the book..)

(Please see the linked figure in post #589.)

Can you tell me how you arrive at the answer? ... ie the equations to use? Why those equations? Also, what does doing postive and negative work mean?


Thanks!

I'll pm you the answer and reasoning.

 

[phospho]

Would someone be so kind enough to help me with this one? 🙂

Thank you in advance😍

Two capacitors are identical, except that one is empty and the other is filled with a dielectric ( = 4.70). The empty capacitor is connected to a 6.0 V battery. What must be the potential difference across the plates of the capacitor filled with a dielectric such that it stores the same amount of electrical energy as the empty capacitor?

 

[phospho]

Would someone be so kind enough to help me with this one? 🙂

Thank you in advance😍

Two capacitors are identical, except that one is empty and the other is filled with a dielectric ( = 4.70). The empty capacitor is connected to a 6.0 V battery. What must be the potential difference across the plates of the capacitor filled with a dielectric such that it stores the same amount of electrical energy as the empty capacitor?

nobody? anybody? please?😳

😍

 

[gridiron]

Would someone be so kind enough to help me with this one? 🙂

Thank you in advance😍

Two capacitors are identical, except that one is empty and the other is filled with a dielectric ( = 4.70). The empty capacitor is connected to a 6.0 V battery. What must be the potential difference across the plates of the capacitor filled with a dielectric such that it stores the same amount of electrical energy as the empty capacitor?

Hey,
You can solve for the potential energy stored by the capacitor hooked up to the battery. The potential energy for the empty capacitor is (.5)CV^2. Now, the presence of the dielectric in the other capacitor weakens the electric field strenght making it easier for more charge to be stored. If you want to find the potential difference that must be applied to the capacitor with the dielectric, with C empty = C with dilectric (the capacitor value), set the value for the potential you obtained earlier to:

U = 0.5kCV^2

Where k is the dielectric constant. You should obtain a smaller value because of the presence of a dielectric. However, had the capacitor been charged, and the porcelain dielectric inserted, the potential energy would decrease---by a factor equal to the dielectric constant. The capacitor would exert a tiny tug on the slab and would do work on it and if the slab were allowed to slide between the plates without friction it would oscillate back and forth with constant mechanical energy equal to the lost potential energy from introduction of the slab. I hope this helps and good luck!!

 

[phospho]

Hey,
You can solve for the potential energy stored by the capacitor hooked up to the battery. The potential energy for the empty capacitor is (.5)CV^2. Now, the presence of the dielectric in the other capacitor weakens the electric field strenght making it easier for more charge to be stored. If you want to find the potential difference that must be applied to the capacitor with the dielectric, with C empty = C with dilectric (the capacitor value), set the value for the potential you obtained earlier to:

U = 0.5kCV^2

Where k is the dielectric constant. You should obtain a smaller value because of the presence of a dielectric. However, had the capacitor been charged, and the porcelain dielectric inserted, the potential energy would decrease---by a factor equal to the dielectric constant. The capacitor would exert a tiny tug on the slab and would do work on it and if the slab were allowed to slide between the plates without friction it would oscillate back and forth with constant mechanical energy equal to the lost potential energy from introduction of the slab. I hope this helps and good luck!!

I just saw your post, thank you so much for taking the time to explain it to me...I got 2.77 Volts and it's the right answer...again, thank you🙂

 

[gridiron]

I have a few questions for future MCAT takers. What topics would you like to see with greater explanation in the physics thread? Would you like more real world applications to problems in order to getter a better grasp of the material? Suggestions are welcome!

 

[superwillis]

Pressure = Force/Area, correct?
I was always under the assumption that the way a bullet penetrates anything is because of the high pressure it applies. But a bullet applies no force, right? The bullet is not accelerating when it hits an object. Therefore, a bullet can't apply a force or apply any pressure to an object it hits.

So, is the reason why bullets can go through objects because its kinetic energy is converted into other forms of energy, like heat, and work (i.e. ripping apart wood, stone, flesh, etc.)?

Is there some term for momentum/area? Because it seems intuitive that the more pointed the bullet, the better it will penetrate an object, so there must be some way for the area over which the energy is exerted can be considered.

Can anyone confirm this?

 

[estairella]

Pressure = Force/Area, correct?
I was always under the assumption that the way a bullet penetrates anything is because of the high pressure it applies. But a bullet applies no force, right? The bullet is not accelerating when it hits an object. Therefore, a bullet can't apply a force or apply any pressure to an object it hits.

F = ma. I shoot your face. The velocity before the bullet hits you is 500m/s. It takes 0.1 s from the point of contact to when its completely stopped by your skull.
v = vo + at
0 = 500 + a(0.1)
a = -5000m/s^2

If a bullet is 10 grams (0.01kg), you're looking at a force of 50N in about 1mm^2 (at the tip of the bullet)... 50N/(0.001m)^2 = 50 MPa (megapascals)!! That's quite a bit of pressure...

Yea I wanted to write a post about shooting someone in the face

Just remember.. decceleration is still acceleration!