# Uniform electrical field question

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#1
A diagram shows two parallel plates that are "oppositely charged", with one being 0V and the other 1V. The plates create a uniform electric field.

The book says the electron experiences a constant electrostatic force and therefore has constant acceleration.

Please may someone explain the following:

1) why does a 0V plate has an opposite charge to a 1V plate?

2) (i think this Q and the next are dependent on that first Q) how is the electrical field strength at the 0V plate equal to the electric field strength at the other plate?

3) (this thought is because of the F = EQ equation) how is the charge on both plates equal in size but opposite in sign if the electric potential on the plates is different?

Thank you to whoever answers this question 🙂, please tell me if I'm not specific enough and at which part 🙂
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#2
(Original post by BobbJo)
1) Question says they are oppositely charged. 1V will be positive since it is at higher potential.
2) Question says the plates create a uniform electric field. Field strength is constant between plates.
3) The plates are connected to a voltage source. One plate is at a higher potential while the other is at a lower potential. The charges are the same but opposite in sign. Try convincing yourself with an explanation similar to here https://www.quora.com/Why-do-the-par...charge-density
Are the plates labelled with their electric potential, for example, will a particle go through 1V of potential difference in travelling from infinity to the plate? 🙂
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#3
(Original post by BobbJo)
Electric potential at a point is the work done in moving unit positive charge from infinity to the point.
What does the V in the 0V represent? 🙂
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#4
(Original post by BobbJo)
volt
A physical quantity is described by the magnitude and unit.
The voltage is being described by 0 and V
Sorry, i meant does it represent potential difference? 🙂
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#5
(Original post by BobbJo)
The plate is at a potential of 0V.
The plate is at an electrical potential of 0V? 🙂
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#6
(Original post by BobbJo)
Yes?
How can the unit be volts when the SI units of electrical potential are kgm²s⁻²A⁻¹ (which is JA⁻¹)? 🙂
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#7
(Original post by BobbJo)
?
It's joule per coloumb?
What's joules per coulomb? 🙂
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#8
(Original post by BobbJo)
1 volt?
Google says electric potential is measured in joules per coulomb but i just simplified the base units to get joules per amp 🙂
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1 year ago
#9
Basically, suppose you have a +5V and a -5V plate in this case the positive plate is the +5V as it is more positive than the -5V. If you have a +3V plate and a -6V plate, +3V is more positive than the -6V plate so it is considered as a positive plate. Finally, if you have a +1V and a 0V plate the +1 is considered more positive so that's the positive plate. However, how do we obtain a positive plate to begin with?- the example with two plates is actually called a capacitor. In a capacitor ( which is basically the two plates connected to a cell/battery) electrons move onto one plate and as a consequence due to electrostatic repulsion the electrons move off the other plate. Hence one plate is positively charged and other is negatively charged. You don't really need to know how this charging process happens when looking at electric fields, I just explained for context. You will look at them after so, it will make sense.
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1 year ago
#10
(Original post by Freedom physics)
A diagram shows two parallel plates that are "oppositely charged", with one being 0V and the other 1V. The plates create a uniform electric field.

The book says the electron experiences a constant electrostatic force and therefore has constant acceleration.

Please may someone explain the following:

1) why does a 0V plate has an opposite charge to a 1V plate?

2) (i think this Q and the next are dependent on that first Q) how is the electrical field strength at the 0V plate equal to the electric field strength at the other plate?

3) (this thought is because of the F = EQ equation) how is the charge on both plates equal in size but opposite in sign if the electric potential on the plates is different?

Thank you to whoever answers this question 🙂, please tell me if I'm not specific enough and at which part 🙂

I think I can understand where are you coming from. I would not answer all the questions one by one because the answers are interlinked.

I would list a few key points/concepts.
• An electric potential value at a point does not have any physical meaning. It is the potential difference that has physical meaning.
• Constant electric field strength does not imply constant electric potential.
• The electric field strength is a measure of the rate of change of the electric potential with respect to position. This implies that the electric field lines always point in the direction of decreasing electric potential.
• Both electric field and potential depend on the source charge distribution.

Consider a pair of uncharged parallel plates connected to a battery of emf 10 V. Each plate is connected to one terminal of a battery, which acts as a source of potential difference. The battery establishes an electric field in the connecting wires when the connections are made.

Let’s focus on the plate connected to the negative terminal of the battery.

The electric field in the wire applies a force on electrons in the wire immediately outside this plate; this force causes the electrons to move onto the plate. The movement continues until the plate, the wire, and the terminal are all at the same electric potential.

Once this equilibrium situation is attained, a potential difference no longer exists between the terminal and the plate; as a result, no electric field is present in the wire and the electrons stop moving. The plate now carries a negative charge.

A similar process occurs at the other plate, where electrons move from the plate to the wire, leaving the plate positively charged.

In the final configuration, the potential difference across the parallel plates is the same as that between the terminals of the battery. The potential difference across the parallel plates is thus 10 V. The charge distribution gives rise to an electric field and electric potential has a value at every point in the electric field.

We can assign the electric potential at the positively and negatively charged plates to be 10 V and 0 V, respectively. However, we can also assign electric potential at the positively and negatively charged plates to be +5 V and -5 V, respectively. There is no difference in the two assignments of the electric potential as both give the same potential difference between the two oppositely charged plates.

The electric field due a charged plate is constant. I am not explaining why the electric field is constant. To understand it, you need to need to learn about Gauss’s law.
http://hyperphysics.phy-astr.gsu.edu...ic/elesht.html

To relate the charges in the plates with the potential difference across the oppositely charged plates and electric field, you need the concept of capacitance.
https://cnx.org/contents/[email protected]
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1 year ago
#11
(Original post by Freedom physics)
How can the unit be volts when the SI units of electrical potential are kgm²s⁻²A⁻¹ (which is JA⁻¹)? 🙂
Are you sure?
https://en.wikipedia.org/wiki/Electric_potential
In terms of the base units, base units of the electric potential is
kg m2 s-3 A-1

(Original post by Freedom physics)
Google says electric potential is measured in joules per coulomb but i just simplified the base units to get joules per amp 🙂
1 V = 1 J/C

How do you simplify J/C to J/A? Are you sure that you have done the right thing?
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1 year ago
#12
(Original post by BobbJo)
…A physical quantity is described by the magnitude and unit.
It seems that you are using a “textbook” or some examination board syllabus definition of the physical quantity. If you are using it to score mark in exam, I think it is ok.

Just a note on this “definition”. The word “magnitude” is IMO a very “strong” word. Because this means that the physical quantity cannot have negative value. I don’t think this is good. There are physical quantities that have negative values.
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#13
(Original post by Eimmanuel)
1 V = 1 J/C

How do you simplify J/C to J/A? Are you sure that you have done the right thing?
I'm sure i made no errors in the simplification of the algebra and units but I don't think i started with the correct equation, i just went with the equation in my formula booklet that says "V = Q / (4π x (epsilon nought) x r²), why wouldn't that work? 🙂
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#14
(Original post by Eimmanuel)
I think I can understand where are you coming from. I would not answer all the questions one by one because the answers are interlinked.

I would list a few key points/concepts.
• An electric potential value at a point does not have any physical meaning. It is the potential difference that has physical meaning.
• Constant electric field strength does not imply constant electric potential.
• The electric field strength is a measure of the rate of change of the electric potential with respect to position. This implies that the electric field lines always point in the direction of decreasing electric potential.
• Both electric field and potential depend on the source charge distribution.

Consider a pair of uncharged parallel plates connected to a battery of emf 10 V. Each plate is connected to one terminal of a battery, which acts as a source of potential difference. The battery establishes an electric field in the connecting wires when the connections are made.

Let’s focus on the plate connected to the negative terminal of the battery.

The electric field in the wire applies a force on electrons in the wire immediately outside this plate; this force causes the electrons to move onto the plate. The movement continues until the plate, the wire, and the terminal are all at the same electric potential.

Once this equilibrium situation is attained, a potential difference no longer exists between the terminal and the plate; as a result, no electric field is present in the wire and the electrons stop moving. The plate now carries a negative charge.

A similar process occurs at the other plate, where electrons move from the plate to the wire, leaving the plate positively charged.

In the final configuration, the potential difference across the parallel plates is the same as that between the terminals of the battery. The potential difference across the parallel plates is thus 10 V. The charge distribution gives rise to an electric field and electric potential has a value at every point in the electric field.

We can assign the electric potential at the positively and negatively charged plates to be 10 V and 0 V, respectively. However, we can also assign electric potential at the positively and negatively charged plates to be +5 V and -5 V, respectively. There is no difference in the two assignments of the electric potential as both give the same potential difference between the two oppositely charged plates.

The electric field due a charged plate is constant. I am not explaining why the electric field is constant. To understand it, you need to need to learn about Gauss’s law.
http://hyperphysics.phy-astr.gsu.edu...ic/elesht.html
Thank you so much for the thorough explanation!! The explanation covered everything but two things: please may you tell me what the"connections" were that you referred to with the uncharged parallel plates connected to the cell of emf 10V and how does the fact that E is a measure of rate of change of electric potential with respect to distance imply field lines always point in the direction of decreasing electric potential? 🙂
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#15
(Original post by BobbJo)
You must have made an error in the simplifcation of units

V = Q / (4π x (epsilon nought) x r²)

V has units As/(A^2s^4kgm^-3 x m^2) =kg m^2 s^−3 A^−1
Yeah it simplifies to joules per coulomb, i have no clue where i went wrong last time lol
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1 year ago
#16
(Original post by Freedom physics)
I'm sure i made no errors in the simplification of the algebra and units but I don't think i started with the correct equation, i just went with the equation in my formula booklet that says "V = Q / (4π x (epsilon nought) x r²), why wouldn't that work? 🙂
The formula for electric potential is incorrect. I would recommend that you check all formula in the booklet. The correct electric potential is
Electric field and electric force follow what is known as inverse square law:
while electric potential and electric potential energy follow:
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1 year ago
#17
(Original post by BobbJo)
You must have made an error in the simplifcation of units

V = Q / (4π x (epsilon nought) x r²)

V has units As/(A^2s^4kgm^-3 x m^2) =kg m^2 s^−3 A^−1
Not sure how do you arrive at the correct SI unit using the wrong formula. 0
1 year ago
#18
(Original post by Freedom physics)
Thank you so much for the thorough explanation!! The explanation covered everything but two things: please may you tell me what the"connections" were that you referred to with the uncharged parallel plates connected to the cell of emf 10V and how does the fact that E is a measure of rate of change of electric potential with respect to distance imply field lines always point in the direction of decreasing electric potential? 🙂
The following website explains how the electric field is related to electric potential. You would need a bit of calculus to appreciate what they are doing.
https://cnx.org/contents/[email protected]

A pre-understanding of the following content is required.
https://cnx.org/contents/[email protected]

I would try to find time to write a short summary to explain it. I believe you can gain a lot of insights by reading the two websites.
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1 year ago
#19
(Original post by Freedom physics)
Thank you so much for the thorough explanation!! The explanation covered everything but two things: please may you tell me what the"connections" were that you referred to with the uncharged parallel plates connected to the cell of emf 10V ….
It is just normal wire that connects the terminals of the battery/cell to to the respective plates.

(Original post by Freedom physics)
….. electric field lines always point in the direction of decreasing electric potential

An easy way is to look this result is to look at Figure 24.28 in the following website.
https://cnx.org/contents/[email protected].

It shows a positive point charge with electric field lines emitting from the center. The dotted concentric circles' lines are equipotential lines.
The electric field lines are pointing in the direction where electric potential values are decreasing.
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1 year ago
#20
(Original post by Freedom physics)
Thank you so much for the thorough explanation!! The explanation covered everything but two things: please may you tell me what the"connections" were that you referred to with the uncharged parallel plates connected to the cell of emf 10V and how does the fact that E is a measure of rate of change of electric potential with respect to distance imply field lines always point in the direction of decreasing electric potential? 🙂
The connection between electric potential and electric field is usually derived in the following form:
Looking at the right-hand side of the equation, we can see that when the infinitesimal displacement vector is pointing in the same direction as that of the electric field, the overall numerical value would tend to be negative which imply that Vb < Va.

However, if the infinitesimal displacement vector is pointing in the opposite direction as that of the electric field, the overall numerical value would tend to be positive which imply that Vb > Va.

Based on these result, we can see the direction of the electric field is pointing “downhill” in the direction of decreasing potential.

To transform the following equation
to
in component form of Cartesian coordinates:
you need multi-variable calculus. Although students tend to learn the derivation using single variable calculus tools, it is shown in the link.

The implication of the formula is that “At each point in an electric field, the potential gradient points in the direction in which V increases most rapidly with a change in position. So at each point, the direction of E is the direction in which V decreases most rapidly and is always perpendicular to the equipotential surface through the point.” This is why E is a measure of the rate of change of electric potential with respect to distance.
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