Double Checking. Electrostatics

[MedPR]

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Ok I have a few questions about this figure.

1. How do you know the top plate of Vb is the positive plate, besides the fact that the cathode ray deflects toward it?


Ok, I guess just one question. Thanks.

 

[milski]

Ok I have a few questions about this figure.

1. How do you know the top plate of Vb is the positive plate, besides the fact that the cathode ray deflects toward it?


Ok, I guess just one question. Thanks.

The battery symbol - it's a convention to have the positive side as a long thin line, negative as a shorter, sometimes thicker one.

 

[MedPR]

The battery symbol - it's a convention to have the positive side as a long thin line, negative as a shorter, sometimes thicker one.

I knew it had something to do with that. This is a convention I've never really understood. I thought the convention was that negative charges move out of the "longer line" and around through the circuit to the "shorter line"?

 

[milski]

I knew it had something to do with that. This is a convention I've never really understood. I thought the convention was that negative charges move out of the "longer line" and around through the circuit to the "shorter line"?

Yes, it's a bit messed up. The convention is that we are talking about positive charges. These are the one that move from positive to negative terminal. In reality the opposite happens - electrons move the opposite way.

 

[MedPR]

Yes, it's a bit messed up. The convention is that we are talking about positive charges. These are the one that move from positive to negative terminal. In reality the opposite happens - electrons move the opposite way.

Ok so the convention for circuits is that positive charges move from the long line to the short line. The long line then is the cathode?

And the reason the plate closer to the longer line is positive is because as positive charges move out of the battery they build up on the plate before they are able to move on through the circuit?

 

[milski]

Ok so the convention for circuits is that positive charges move from the long line to the short line. The long line then is the cathode?

And the reason the plate closer to the longer line is positive is because as positive charges move out of the battery they build up on the plate before they are able to move on through the circuit?

Yes and yes.

 

[SaintJude]

MedPr, what's the answer ?! I drew the force, than an electron would experience in each field within each circuit.

i said increasing Vb only.

 

[BerkReviewTeach]

That is an excellent thought-inducing question. It is conceptual and makes you look at a complicated device in a simple manner, which is exactly what you need to do on the MCAT. Whoever wrote it did a fantastic job. Where is it from?

 

[chiddler]

MedPr, what's the answer ?! I drew the force, than an electron would experience in each field within each circuit.

i said increasing Vb only.

it's C. Increasing the speed at which the electron is shot increases the force from the Bfield on the electron and makes it go opposite of the Efield = lower. Also, increasing Bfield strength will also push it downwards.

Berkreviewteach. Here.

 

[hellocubed]

increasing Vb

This would only increase the curvature of the electron, from a greater buildup of positive charge on the top plate.

 

[SaintJude]

it's C. Increasing the speed at which the electron is shot increases the force from the Bfield on the electron and makes it go opposite of the Efield = lower. Also, increasing Bfield strength will also push it downwards.

Berkreviewteach. Here.

About Va: The force felt due to the electric field on the electron is antiparallel. It's "straight" in the first place. So I don't see how increasing the acceleration would "straighten" the path.

Ok, the electric field created by Vc induces an upward force on the electron, doesn't it. So an increase in Vc, would lead to a greater force deflection upward. So how does that "straighten the path"

 

[MedPR]

MedPr, what's the answer ?! I drew the force, than an electron would experience in each field within each circuit.

i said increasing Vb only.

That is an excellent thought-inducing question. It is conceptual and makes you look at a complicated device in a simple manner, which is exactly what you need to do on the MCAT. Whoever wrote it did a fantastic job. Where is it from?

it's C. Increasing the speed at which the electron is shot increases the force from the Bfield on the electron and makes it go opposite of the Efield = lower. Also, increasing Bfield strength will also push it downwards.

Berkreviewteach. Here.

About Va: The force felt due to the electric field on the electron is antiparallel. It's "straight" in the first place. So I don't see how increasing the acceleration would "straighten" the path.

Ok, the electric field created by Vc induces an upward force on the electron, doesn't it. So an increase in Vc, would lead to a greater force deflection upward. So how does that "straighten the path"

Sorry guys, I didn't realize I didn't post the answer.

BRT it is from wikipremed.com lectures.

There are actually two answers. I believe it was A and D. I can't see the image right now because imgur is blocked at work. If someone wants to reattach it I will tell you the answers (and explanation).

 

[chiddler]

About Va: The force felt due to the electric field on the electron is antiparallel. It's "straight" in the first place. So I don't see how increasing the acceleration would "straighten" the path.

Ok, the electric field created by Vc induces an upward force on the electron, doesn't it. So an increase in Vc, would lead to a greater force deflection upward. So how does that "straighten the path"

you should look at magnetic field force instead of electric field equation. Fb = qvB. More velocity = more force! Since Efield is pulling up and Bfield down, more velocity will cause a stronger force from the magnetic field and thus more down force.

second, Vc will increase the strength of the magnetic field (B in the Fb equation above) so both cause stronger force downwards by increasing the strength of the mag field.

 

[MedPR]

Just looked at wikipremed. I and IV are the answers. So answer C.

 

[SaintJude]

Ok, yeah, so it doesn't have much to do with Vc, I understand that...but what's Va got to do with this? The electric field due to Va is parallel, already in direction of the motion of the desired path! Increase velocity does not "straighten the path"...

 

[MedPR]

Ok, yeah, so it doesn't have much to do with Vc, I understand that...but what's Va got to do with this? The electric field due to Va is parallel, already in direction of the motion of the desired path! Increase velocity does not "straighten the path"...

Once the particle is in the magnetic field, it is subject to the mag field force.

The equation to look at is F=qvB. If you increase the velocity of a particle moving in a mag field, you also increase the force exerted by the mag field.

The question tells you that the mag force opposes the electric field force. Since the cathode ray deflects up, the force of the e-field on the particle is upward, so the mag field must point downward. So, increasing the mag field will push the particle down = straighten it out.

 

[MedPR]

Ok, yeah, so it doesn't have much to do with Vc, I understand that...but what's Va got to do with this? The electric field due to Va is parallel, already in direction of the motion of the desired path! Increase velocity does not "straighten the path"...

And it does have to do with Vc. If you increase the voltage, you increase the current, and increasing the current increases the force of the field.

 

[SaintJude]

Wait, a minute, MedPr, the "force due to the electric field" is pointing from left to right, isn't it? I really, really think-almost 90% sure it is. Just by the designation of the battery potential, you can tell. It's not pointing "upward"

 

[MedPR]

Wait, a minute, MedPr, the "force due to the electric field" is pointing from left to right, isn't it? I really, really think-almost 90% sure it is. Just by the designation of the battery potential, you can tell. It's not pointing "upward"

When it leaves initially the field is pointing right to left (remember, e-field points in the direction protons will move, not electrons), but once it gets under the plate suplied by Vc, it is pointing down.

 

[SaintJude]

Dude, your explanation is suddenly in favor of Vc, as chiddler originally explained not Va as you indicate. I'm ok with letting this question die...

 

[MedPR]

Dude, your explanation is suddenly in favor of Vc, as chiddler originally explained not Va as you indicate. I'm ok with letting this question die...

My explanation is in favor of both, because increasing either of them have the effect of straightening the ray.

 

[SaintJude]

Yeah, but then C isn't the answer. It's A & D, like you said-those choices don't include Vc. I don't think this questions is as well modeled as originally credited. I think the Kaplan question I sent you earlier via message is testing the same thing. Thanks for putting up with me, though.

 

[MedPR]

Yeah, but then C isn't the answer. It's A & D, like you said-those choices don't include Vc. I don't think this questions is as well modeled as originally credited. I think the Kaplan question I sent you earlier via message is testing the same thing. Thanks for putting up with me, though.

By A and D I meant "I" and "IV" I couldn't see the picture at the time and I forgot it was a roman numeral question.

The answer is C, I just looked 🙂 I promise.

Chiddler's original explanation is a good one, and my previous explanation is basically word for word of what the guy says on wikipremed.