Two equation for voltage don't seem to be linking

[September24]

Before I start, I wanna apologize for the confusing nature of my question. I'm so confused that even my question is confusing. Haha In simple, voltage is confusing me and I was wondering if someone can explain me the formulas for voltage qualitatively and how they make sense.


Formula for v=ed and v=kQ/r

To make it simple, we can have a uniform electric field where electric field is the same.

In a scenario, where I have a positive point charge making an electric field, we have voltages that are created as a result of this field. However, how is it that from the first equation, as distance increAses, voltage increases, but in the second, voltage decreases with distance.


Can someone explain what is meant by distance or radius in each equation. Is it distance from the point charge or radius of the test charge... I don't know. Voltage is a property of the electric field and not the test charge so it makes sense that if we are closer to the point charge, the voltage is high but v=Ed doesn't explain this.


If someone can give a qualitative definition of voltage in comparison to the field or something, it may help. Voltage confuses me in general

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[Paean1]

First of all, lets say d = r. I'll just use d instead of r for everything to make it simpler. This variable refers to the distance between the source charge creating the electric field and the point charge affected by it.

Lets look at V=Ed. We also know E=(kQq)/d^2. So, as d increases E decreases by a squared factor (if d is twice as large, E is 1/4th as large). Thus in V=Ed, if you increase d, V will go down because E is decreased by a squared factor (so we may have V=(E/4) * 2d = (Ed)/2).

A key point to understand is that electric field strength decreases as you travel farther from the source charge that is creating the electric field.

Therefore there is no discrepancy between the two equations for voltage. In both cases voltage will go down by the same factor as distance increases.

 

[September24]

Hey again. Looks like you've been my electrostatics mentor lately. Haha

My original questions seems to click. When using the V=Ed question, the concept reminds me of the ideal gas law. Pv=nRT. If we double volume, pressure doesn't automatically half because we don't know of temperature or moles are held constant. Similar, doubling the distance in V=Ed doesn't automatically double the voltage since the E field isn't held constant. Distance doubles but according E=kQ/r^2, electric field is cut to a quarter of its value. So doubling distance and reducing electric field to one fourth has a net value of cutting the voltage to half.
This makes sense since we know the field should decrease as we move away from the source. With voltage being related to the field, voltage should decrease too.


I wanna see if you can apply the same concept to this question fromBerkeley review.

"A charged particle is suspended between two horizontally aligned parallel plates carrying opposite charges of equal magnitudes. If the separation between the plates is doubled while voltage is held constant, what happens to electric force on a particular drop."

The answer is that it halves. Berkely explained that if voltage is constant and v=Ed, double distance should half the field. By F=Eq, half the field and we half the force. This explanation makes sense but once again I go back to e=kQ/r^2. Doubling the distance halved the field instead of cutting it to a quarter. I guess this is because voltage was constant but it's so tricky.

I'm mainly worried since in fast paced nature of the MCAT, I would be quick to jump to one equation and mess up. Is there a thought process you use to evaluate these types of questions? When is it safe to use the F=kQq/r^2 of e=kq/r^2 equation when the question asks me simply what happens to the field or force.

 

[Paean1]

Yes, E halves because voltage is constant. In order for voltage to stay constant when doubling distance, the Q in E=kQ/r^2 has to increase (double), which is why E does not become 1/4th. Although the problem does not state that Q increases once the plates are pulled apart, it must be assumed so if voltage is to remain constant.

When thinking through these problems you should never assume a variable remains constant unless stated that it does. What I've found helpful is to get an intuitive idea (if possible) of what should happen, then begin to apply equations while being very careful about how you manipulate the equations and the assumptions you make. It just takes practice. The MCAT is a very detail oriented exam.

 

[osckey]

We also know E=(kQq)/d^2.

That's not electric field, that's the equation for electrostatic force.

 

[dudewheresmymd]

Hey again. Looks like you've been my electrostatics mentor lately. Haha

I wanna see if you can apply the same concept to this question fromBerkeley review.

"A charged particle is suspended between two horizontally aligned parallel plates carrying opposite charges of equal magnitudes. If the separation between the plates is doubled while voltage is held constant, what happens to electric force on a particular drop."

The answer is that it halves. Berkely explained that if voltage is constant and v=Ed, double distance should half the field. By F=Eq, half the field and we half the force. This explanation makes sense but once again I go back to e=kQ/r^2. Doubling the distance halved the field instead of cutting it to a quarter. I guess this is because voltage was constant but it's so tricky.

I'm mainly worried since in fast paced nature of the MCAT, I would be quick to jump to one equation and mess up. Is there a thought process you use to evaluate these types of questions? When is it safe to use the F=kQq/r^2 of e=kq/r^2 equation when the question asks me simply what happens to the field or force.

you have it mixed up. KQ/r^2 is used for the electric field of a point charge. See below. You use the V=ED when the electric field is assumed to be constant between two plates that have large areas as compared to the distance between them. Thus, regardless of where a test charge is placed between two plates, the magnitude and direction of the electric force (Fe) produced will be the same (Ie. the force acting on a charged particle between two parallel plates such as those in a capacitor is constant)

When you see the words parallel plates you use V=ED, not E=kQ/r^2. Therefore, if voltage is constant, and d is doubled, E must be halved. Do not use KQ/r^2 when looking at parallel plates.

 

[September24]

you have it mixed up. KQ/r^2 is used for the electric field of a point charge. See below. You use the V=ED when the electric field is assumed to be constant between two plates that have large areas as compared to the distance between them. Thus, regardless of where a test charge is placed between two plates, the magnitude and direction of the electric force (Fe) produced will be the same (Ie. the force acting on a charged particle between two parallel plates such as those in a capacitor is constant)

When you see the words parallel plates you use V=ED, not E=kQ/r^2. Therefore, if voltage is constant, and d is doubled, E must be halved. Do not use KQ/r^2 when looking at parallel plates.


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Thanks for the help. Could you answer this question about uniform electric fields?


Uniform electric field.

I'm copying a question from another sdn post.

"I came across this question:

What change is observed in the force exerted on charged particle P in a uniform electric field when it is moved closer to the plate on the left, so that the distance r separating them is reduced by half?

A. The force doubles, if the charge is negative.
B. The force doubles, if the charge is positive.
C. The force increases by a factor of 4, regardless of the type of charge.
D. The force does not change, regardless of the magnitude or sign of the charge."

The answer was D here. I get it since the electric field is the same and doesn't change. Since the E field the same throughout the area between the plates, force does not change according to F=Eq.



Then I found this question.

"A charges particle is suspended between two horizontally parallel plates carrying charges of equal magnitude. If the separation between the plates doubled while voltage is held constant, what happens to force on a particular drop".

The answer was "it is halved" but I put that it is the same. V=Ed. If voltage is held constant, increasing distance would make the field decrease. F=Eq. Decrease the field, force decreases. Mathematically, this makes sense but I thought that changing distance wouldn't change force. What's changing here?

 

[dudewheresmymd]

Thanks for the help. Could you answer this question about uniform electric fields?


Uniform electric field.

I'm copying a question from another sdn post.

"I came across this question:

What change is observed in the force exerted on charged particle P in a uniform electric field when it is moved closer to the plate on the left, so that the distance r separating them is reduced by half?

A. The force doubles, if the charge is negative.
B. The force doubles, if the charge is positive.
C. The force increases by a factor of 4, regardless of the type of charge.
D. The force does not change, regardless of the magnitude or sign of the charge."

The answer was D here. I get it since the electric field is the same and doesn't change. Since the E field the same throughout the area between the plates, force does not change according to F=Eq.



Then I found this question.

"A charges particle is suspended between two horizontally parallel plates carrying charges of equal magnitude. If the separation between the plates doubled while voltage is held constant, what happens to force on a particular drop".

The answer was "it is halved" but I put that it is the same. V=Ed. If voltage is held constant, increasing distance would make the field decrease. F=Eq. Decrease the field, force decreases. Mathematically, this makes sense but I thought that changing distance wouldn't change force. What's changing here?

You're reasoning is correct but I believe you are memorizing the equations without understanding what they are telling you. I bolded/underlined what's important to understand for emphasis.

1st case-->THE DISTANCE BETWEEN THE PLATES IS CONSTANT (D), THEREFORE ELECTRIC FIELD (E) IS CONSTANT NO MATTER WHERE YOU PLACE THE TEST CHARGE. Force on a test charge =qE , b/c E doesn't change, the force is constant whether the charge is 1cm from the left plate, 10 cm from the right plate, 2 cm from the left plate, 4 cm from the right plate, it doesn't matter! What's changing is the distance between the test charge and the plate which is NOT the same as the distance between the two plates.

2nd Case--> THE DISTANCE BETWEEN THE PLATES (D) IS CHANGING THEREFORE ELECTRIC FIELD STRENGTH CHANGES (E). Let's apply similar reasoning from the first case. We are dealing with two parallel plates. Therefore, the electric field equation for the strength of the electric field is V/D=E (remember not to use E field = kQ/r^2 as we are talking about plates and not point charges). Now that we have established that Electric field between the two plates is now in units 0f (VOLTS/METER) question 2 states that the separation of the plates is doubled, while voltage is held constant. If so, Because E=V/D, if we double the separation of the plates (d) then we are going to halve the electric field. That is the strength of the electric field in (Volts/Meter) between the two plates is going to decrease by a factor of 2. Now that we know what happens to the strength of the electric field we use the equation F=qE.

Force on a test charge within an electric field (Fe) = (q charge that is being examined) * (Electric field )
We just established above that the electric field strength between two plates is going to decrease by a factor of 2, and because the charge on the drop is not changing, the force on the test charge due to the electric field (Fe) must decrease by a factor of 2 everywhere between the two plates..thus it is "halved".

Keep the distance between the plates separate from the distance between the charge and the plate. Another question they might ask is what is the acceleration of an electron between two charged plates? Because force on the charge is constant between the two plates and the mass of the electron isn't changing, an electron in a uniform electric field between two parallel charged plates experiences constant acceleration.

 

[Paean1]

Thanks for explaining this, I was confusing the plates vs. point charges formulas.

 

[dudewheresmymd]

Thanks for explaining this, I was confusing the plates vs. point charges formulas.

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