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 05-01-2012, 04:01 PM #1 Senior Member   Join Date: Feb 2012 Posts: 116 electricity capacitors SDN Members don't see this ad. (About Ads) -------------------- (-) -------------------- (+) 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?...
 05-01-2012, 06:52 PM #2 1K Member   Status: Pre-Medical Join Date: Jan 2012 Posts: 1,479 cloak25, there have been 123 views and no answer--which main equations does the explanation say to use?
 05-01-2012, 06:59 PM #3 1K Member     Status: Pre-Medical Join Date: Dec 2009 Location: Where the rain grows Posts: 1,833 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.
05-01-2012, 07:01 PM   #4
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Quote:
 Originally Posted by 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.

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?

05-01-2012, 07:11 PM   #5
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Quote:
 Originally Posted by 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.
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

05-01-2012, 07:26 PM   #6
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Quote:
 Originally Posted by chiddler 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..

 05-01-2012, 07:33 PM #7 2K Member   Join Date: Apr 2010 Posts: 2,388 ty. tried to do that initially i couldn't think how to relate distance to that >.>
 05-01-2012, 07:36 PM #8 1K Member     Status: Pre-Medical Join Date: Dec 2009 Location: Where the rain grows Posts: 1,833 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.
 05-01-2012, 08:28 PM #9 Senior Member   Join Date: Feb 2012 Posts: 116 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
05-01-2012, 08:56 PM   #10
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Quote:
 Originally Posted by 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
that's what i did but is incorrect because this solution makes d into a square root. none of the answer choices reflect that.

05-01-2012, 09:07 PM   #11
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Quote:
 Originally Posted by SaintJude Why does a higher dielectric constant mean that particle has a harder time moving?

"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

05-01-2012, 09:17 PM   #12
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Quote:
 Originally Posted by pm1 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.

 05-01-2012, 09:24 PM #13 1K Member     Status: Pre-Medical Join Date: Dec 2009 Location: Where the rain grows Posts: 1,833 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.
 05-01-2012, 09:38 PM #14 2K Member   Join Date: Apr 2010 Posts: 2,388 thanks a very good correction.
 05-03-2012, 10:48 PM #15 Member   Join Date: Mar 2009 Posts: 31 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)
05-03-2012, 10:55 PM   #16
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Quote:
 Originally Posted by 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.
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