electricity capacitors

[cloak25]

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So drawn above is a parallel plate capacitor separated by a distance "d" .
A proton (mass m) is placed on top of the positively charged plate. The charge on the capacitor is Q and the capacitance is C. If the E-field in the region between the plates has a magnitude of E, which of the following expressions gives the time required for the proton to move up the other plate?

A. d√(eQ/mC)
B. d√(m/eQC)
C. d√(2eQ/mC)
D. d√(2mC/eQ)

answer: D this might be a good mcat question?...

 

[SaintJude]

cloak25, there have been 123 views and no answer--which main equations does the explanation say to use?

 

[milski]

Ignoring equations for a moment:
- higher mass will make the particle move slower and take more time -> mass has to be in the enumerator
- higher capacitance would imply a higher dielectric constant, harder for the particle to move -> capacitance has to be in the denominator.
done.

You can also play around with the units but that's a bit trickier and requires you to remember more things.

 

[SaintJude]

Ignoring equations for a moment:
- higher mass will make the particle move slower and take more time -> mass has to be in the enumerator
- higher capacitance would imply a higher dielectric constant, harder for the particle to move -> capacitance has to be in the denominator.
done.

You can also play around with the units but that's a bit trickier and requires you to remember more things.

😱 milski you're brilliant. I'm going to keep that mind---I never thought of determining the answer by using the physical relationships...lol--i'm a bit slow 😳

Why does a higher dielectric constant mean that particle has a harder time moving?

 

[chiddler]

Ignoring equations for a moment:
- higher mass will make the particle move slower and take more time -> mass has to be in the enumerator
- higher capacitance would imply a higher dielectric constant, harder for the particle to move -> capacitance has to be in the denominator.
done.

You can also play around with the units but that's a bit trickier and requires you to remember more things.

i figured as much, but can you think of a way to use equations? i got t = (2dm/qE)^0.5..i can't think of a way to keep d outside the square root.

i was using d = vo*t + 1/2 at^2 and F=qE

 

[pm1]

i figured as much, but can you think of a way to use equations? i got t = (2dm/qE)^0.5..i can't think of a way to keep d outside the square root.

i was using d = vo*t + 1/2 at^2 and F=qE

I did it like this..

C=Q/V, then V=Q/C

If we want to find the velocity in order to determine the time we probably want to use kinetic energy, which we can get from potential energy. U=kqq/r so V=U/q since the potential energy is going to have turned into KE, then we have V=mv^2/2q

mv^2/2q = Q/C (q=e)
v^2 = 2eQ/mC
v= sqrt 2eQ/mC

since v=d/t --> t=d/v

then we get
t= d*sqrt mc/2eQ

I just don't get why my 2 is on the denominator and not in the numerator like the answer choice..

 

[chiddler]

ty.

tried to do that initially

i couldn't think how to relate distance to that >.>

 

[milski]

On the phone, so rather brief:

Find the final velocity when the particle reaches the other plate from energy conservation (potential turning into kinetic)

Keeping in mind that the motion is with uniform acceleration, the average speed will be half of the final speed.

Calculate time based in distance and average speed.

I have not tried this but it should work.

 

[cloak25]

explanation says to find acceleration first. once you find acceleration, you can use y=1/2at^2 to find time.

F=ma
a = F/m
a = qE/m = eE/m.
E=V/d and V=Q/C
so, a = eQ/mdC. substituting eQ/mdC for acceleration into d=1/2at^2

answer D.

this is how you solve it using equations, obviously but I just wanted to see if anyone can explain it using physical relationships to gain a better understanding conceptually. milski came the closest though 🙂

 

[chiddler]

explanation says to find acceleration first. once you find acceleration, you can use y=1/2at^2 to find time.

F=ma
a = F/m
a = qE/m = eE/m.
E=V/d and V=Q/C
so, a = eQ/mdC. substituting eQ/mdC for acceleration into d=1/2at^2

answer D.

this is how you solve it using equations, obviously but I just wanted to see if anyone can explain it using physical relationships to gain a better understanding conceptually. milski came the closest though 🙂

that's what i did but is incorrect because this solution makes d into a square root. none of the answer choices reflect that.

 

[SaintJude]

Why does a higher dielectric constant mean that particle has a harder time moving?

Found answer to my question:

"At first, it is easy to push charge on to the capacitor, as there is no charge there to repel it. As the charge stored increases, there is more repulsion and it is harder (more work must be done) to push the next lot of charge on." -IOP

 

[milski]

I did it like this..

C=Q/V, then V=Q/C

If we want to find the velocity in order to determine the time we probably want to use kinetic energy, which we can get from potential energy. U=kqq/r so V=U/q since the potential energy is going to have turned into KE, then we have V=mv^2/2q

mv^2/2q = Q/C (q=e)
v^2 = 2eQ/mC
v= sqrt 2eQ/mC

since v=d/t --> t=d/v

then we get
t= d*sqrt mc/2eQ

I just don't get why my 2 is on the denominator and not in the numerator like the answer choice..

What you have found is the final velocity - remember, it's a uniformly accelerating motion. The average velocity is half of that and the time will be twice what you've found, which will bring the 2 in the top part under the square root.

 

[milski]

Both energy and kinetics solutions work.
pm1 already did the energy solution, the only correction there is the switch from final velocity to average velocity.
cloak25 has the correct kinetics solution. chiddler, there are two d-s multiplied under the square root, one comes from t=sqrt(2d/a), the other comes from a=eQ/mdC. That gives you a d in front of the radical.

 

[chiddler]

thanks a very good correction.

 

[podyon18]

Given: q,d,Q,C,E

d = 1/2 at^2
t^2 = 2d/a

F= m/a
F = eE (E = d/V) (Q = CV) (E = Q/dC) so
F = eQ/dC = ma

a = eQ/dCm

t^2 = 2d/a
t^2 = 2Cmd^2 /eQ

take square root of both sides and.......
t = d√(2mC/eQ)

 

[Morsetlis]

Ignoring equations for a moment:
- higher mass will make the particle move slower and take more time -> mass has to be in the enumerator
- higher capacitance would imply a higher dielectric constant, harder for the particle to move -> capacitance has to be in the denominator.
done.

You can also play around with the units but that's a bit trickier and requires you to remember more things.

Your brilliance shall echo through the ages.

 

[sshah92]

Hey guys,

Also reviewing this and wanted to go over the limiting variables strategy that milski used.

mass: higher mass indicates the proton takes more time to traverse the space; variable must be in numerator.
charge (Q): higher charge indicates proton is more repelled by positive plate and attracted by negative plate; takes less time to traverse space, so it must be in the denominator

This leaves us with choices B and D. Both have m in numerator and Q in denominator. I could go through the algebra but the MCAT isn't designed for us to spend so much time per question, so I'd rather figure out the intuition behind capacitance. Why is capacitance in the numerator?

The answer was D. d√(2mC/eQ)

 

[sshah92]

Hey guys,

Also reviewing this and wanted to go over the limiting variables strategy that milski used.

mass: higher mass indicates the proton takes more time to traverse the space; variable must be in numerator.
charge (Q): higher charge indicates proton is more repelled by positive plate and attracted by negative plate; takes less time to traverse space, so it must be in the denominator

This leaves us with choices B and D. Both have m in numerator and Q in denominator. I could go through the algebra but the MCAT isn't designed for us to spend so much time per question, so I'd rather figure out the intuition behind capacitance. Why is capacitance in the numerator?

The answer was D. d√(2mC/eQ)

Bump! Anybody know why capacitance was in the numerator?

 

[Romz]

Bump! Anybody know why capacitance was in the numerator?

Wow... just when I thought I had E&M down....

My solving:


1/2 mv^2 = 1/2 CV^2

v = √(CV^2/m)

v=d/t = our expression. Lets isolate for t-->

t = d / √(CV^2/m)

t = d√(m/cV^2)

t= d√(m/(q/V)V^2)

t = d√(m/QV)

t = d√(mC/Qq) where q = e

t = d√(mC/Qe)

I am missing the 2 in the denominator though. it cancelled out at the beggining. I know the 1/2 in both equations comes from the integration of W equations, but I still can't figure out why I am missing my two at the top.

 

[sshah92]

Still confused by this and just got out of E&M.

Wouldn't higher capacitance mean there is more charge stored on the plates? Ergo more negative charges on the top plate and more positive charges on the negative plate, so a positive charge resting on the bottom would, given increased capacitance, take LESS time to move to the top plate?! I understand the algebraic solutions (both kinetics and energy) but don't understand why intuition is failing me here

 

[jetsfan1234]

Still confused by this and just got out of E&M.

Wouldn't higher capacitance mean there is more charge stored on the plates? Ergo more negative charges on the top plate and more positive charges on the negative plate, so a positive charge resting on the bottom would, given increased capacitance, take LESS time to move to the top plate?! I understand the algebraic solutions (both kinetics and energy) but don't understand why intuition is failing me here

No, the voltage is the "force" that pushes the particle to move. Higher capacitance just means that more charge is stored for a given voltage difference between the plates.

 

[sshah92]

No, the voltage is the "force" that pushes the particle to move. Higher capacitance just means that more charge is stored for a given voltage difference between the plates.

Voltage causes an electron to move in a circuit from one terminal to another. Potential difference causes the particle (outside of any connecting wire between both plates) to move from one plate to the other. Both are in the same units but I'd be careful about making that assumption.

But back to my question -- If there's a higher capacitance, or more charge stored on the plates, then it takes less "force" for the positively-charged particle to move to the negative plate. So by intuition, you would think that C would be in the denominator.

Sooo why is it in the numerator?

 

[jetsfan1234]

Voltage causes an electron to move in a circuit from one terminal to another. Potential difference causes the particle (outside of any connecting wire between both plates) to move from one plate to the other. Both are in the same units but I'd be careful about making that assumption.

But back to my question -- If there's a higher capacitance, or more charge stored on the plates, then it takes less "force" for the positively-charged particle to move to the negative plate. So by intuition, you would think that C would be in the denominator.

Sooo why is it in the numerator?

To be more specific, voltage is the electric potential energy per unit charge. It has nothing to do with whether you are measuring along a wire or between plates.

Think of it like gravitational potential energy. It's essentially the same concept. A particle travels from high to low voltage (due to the electric force, or whatever you want to call it) just like an apple goes from high--> low potential energy as it falls from a tree.

The acceleration of the particle is not dependent on capacitance, it is dependent on the voltage, which is defined as V = Q/C. If distance = 1/2at^2, You can see that C must be in the numerator with some quick algebra.