AAMC 9 #2 & #4 (electricity)

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IRASNA

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Here are two questions relating to a particle and E Field/Force. I've attached pictures of the two questions.

For #2: I thought the force would increase by a factor of two because F=ma...a 2x change in velocity would mean there is acceleration...so it would be 2F, right? The explanation given by AAMC and Kaplan just mentions F=Eq, so then I thought my logic must be wrong and I shouldn't relating F=ma to F=Eq, until...

I get to #4 and I notice that the question has me relating F=ma=Eq....

Can someone explain why I'm not thinking about this right? Thanks!
 

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Acceleration (#4) isn't the same as speed (#2). Acceleration is actually the derivative of speed, or velocity.

You're quite right that Fe = Eq = ma.
 
So I can't say that doubling velocity means I've doubled acceleration? In my head it made sense at the time I guess
 
Here are two questions relating to a particle and E Field/Force. I've attached pictures of the two questions.

For #2: I thought the force would increase by a factor of two because F=ma...a 2x change in velocity would mean there is acceleration...so it would be 2F, right? The explanation given by AAMC and Kaplan just mentions F=Eq, so then I thought my logic must be wrong and I shouldn't relating F=ma to F=Eq, until...

I get to #4 and I notice that the question has me relating F=ma=Eq....

Can someone explain why I'm not thinking about this right? Thanks!

For #2: F=ma, a=deltaV/deltat[dt]
In order for the force to change, there has to be a change in velocity OVER CHANGE IN TIME. The questions is worded in a way that implies two separate events. In the first event, a particle travels at a certain speed. Then the particle is traveling at twice the speed. The particle itself is not changing speed in time. They simply ask how the change in velocity affects the force...and it doesn't. If you are traveling at a constant speed of V and then at a constant speed of 2V, the net force is still equal to zero because your speed is constant and not changing in time. acceleration is the rate of change of velocity. Since a = 0, F=0 both times.
 
Here are two questions relating to a particle and E Field/Force. I've attached pictures of the two questions.

For #2: I thought the force would increase by a factor of two because F=ma...a 2x change in velocity would mean there is acceleration...so it would be 2F, right? The explanation given by AAMC and Kaplan just mentions F=Eq, so then I thought my logic must be wrong and I shouldn't relating F=ma to F=Eq, until...

I get to #4 and I notice that the question has me relating F=ma=Eq....

Can someone explain why I'm not thinking about this right? Thanks!

For #4:
mass=m
charge=q
parallel plate capacitor.So the E field lines point from positive plate to negative plate. and the E is uniform throughout.

Midway between the plates, the positively charged q will be accelerated toward the negatively charged plate. The acceleration occurs because a force is felt by the charge that is equal to: F=qE.
The acceleration can be calculated then as F=qE=ma.
Thus: qE=ma
(qE)/m = a
Think of it this way. All matter has mass. Velocity is change in position/ change in time. Acceleration in change in velocity/ change in time. (really they are integrals and derivatives of the respective functions with respect to time but for mcat purpose this isn't really necessary) An object can move at a constant velocity. If this is so, there is no acceleration. If an object is not accelerating, then a=0 and F=0. IF you apply a force to an object, you will cause its velocity to either increase or decrease IN TIME. Meaning that it will either speed up per second or slow down per second. In order for there to exist a force, there must be an acceleration and mass(except when you talk about the force applied by electromagnetic radiation that travels at a constant speed c and is massless). So in problem number two, the electron isn't speeding up in time. The rate of change of the velocity is 0 and it just asks you info about two separate events in which the particle travels at two different speeds.
 
So I can't say that doubling velocity means I've doubled acceleration? In my head it made sense at the time I guess

No you can not. Now if the question said that the velocity doubled in a time of 2.3 seconds, then you can say that there is an acceleration. But if the velocity isn't changing in time, there is no acceleration at all.
 
For #2, I just used the electric force equation Ef = (K)(Q1)(Q2)/(R^2), hopefully you're familiar with this. The equation says nothing about velocity or even acceleration. Only things that matter are charge and radius.
 
For #2, I just used the electric force equation Ef = (K)(Q1)(Q2)/(R^2), hopefully you're familiar with this. The equation says nothing about velocity or even acceleration. Only things that matter are charge and radius.

Hahah that too is a way to look at it. Simpler than the way I explained. Really it is all the same thing because f=kq1q2/r^2 and E=f/q so kq1/r^2=E when determining E at point p that has a test charge of q0 which is very small in comparison to q1! So you could even go so far as to say that ma/q= kq/r^2 and solve for a. Then a equals dv/dt which can only occur when dr/dt exists and F is non zero.
🙂 physics is so much fun
 
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