# Find the E-Field between plates when given only given part of the total voltage

#### mhrob24

Homework Statement
Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. What is the magnitude of the electric field strength between them, if the potential 7.85 cm from the zero volt plate (and 2.15 cm from the other) is 693 V?
Homework Equations
E = V/D
I'm really unsure about how to solve this because I am not given the total voltage between the plates. The voltage given (693V) is at a point that is 7.85 cm away from the zero volt plate. If I was given the total voltage between the complete distance between the plates (10 cm), then I could just use E=V/D and solve. But this is not the case. My initial thought was to just say that the total distance between the plates is 7.85 cm, not 10cm. Then, I can solve for the E-field using the voltage given and the distance of 7.85 cm:

E = 693/.0785m = 8828.025

Then, I could find the voltage between the plates at 10cm separation using this value as my E-field, then use this voltage to solve for my E-field:

V = E*(.1m) = 8828.025*(.1m)

However, this makes no sense because the E field that I calculated was for a voltage of 693 and distance of 7.85cm between the plates. If E=V/D, then the E field will change in magnitude as the distance between the plates changes. So I cant just plug in this value of E in order to find the voltage between the plates at 10cm apart. My brain is just hurting now. Please help.

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#### gleem

Step back and take a deep breath. So what is the definition of an electric field?

#### gleem

Actually it is better to determine the nature of the electric field between two parallel plate.

#### mhrob24

Assuming that they both have same magnitude charge and opposite signs, then the electric field will be uniform; extending from the positive plate to the negative plate.

#### gleem

IF the electric field is constant what does that imply about the potential difference?

#### mhrob24

I don't think this is the answer you're looking for but it's the only thing I can think of at the moment:

That V = E*D so the potential difference would also be constant (as long as the distance stays the same)?
If that's the case though, then that would mean I could still use E=V/D with my V at 693 because it is constant for a constant E-field. However, that means E = 693/.1m = 6930 v/m. I entered that in for my answer (this question is from an online HW platform), and that is also an incorrect answer.

(PS thank you for the quick response.)

#### gleem

That V = E*D so the potential difference would also be constant
Stop. The potential difference is constant for any distance. The 10 cm separation is a distraction and irrelevant.
Are you getting my drift. It is late here and I am tired. Talk to you tomorrow.

#### kuruman

Homework Helper
Gold Member
You know that the potential (not the potential difference) as a function of distance $x$ from the zero-potential plate is given by $V(x)=E~x$. You also know that $V(7.85~\mathrm{cm}) = 693 ~\mathrm{Volts}$. Can you find $E$?

#### gleem

The potential difference is constant for any distance.
Definitely was tired. This statement is not correct . It should have read the ratio of potential difference for any corresponding distance is constant. The electric field between two parallel charged plates is constant in the distance between the plates.

#### mhrob24

Good morning. I think I understand now but I want to make sure ...(getting the right answer without understanding isn't going to help me)

So, in this case we have the plates at 10 cm apart. So when they say that the potential difference at 7.85 cm away from the zero volt plate is 693v, that is just trying to confuse the reader because as long as the plates are held at 10cm apart, then regardless of where you're at in between the plates, the potential difference will remain 693v (as long as the field is assumed to be constant). So then the problem becomes very easy:

E = 693v/.1m = 6930 v/m

....I know this was supposed to be a simple question but I over thought it

#### gleem

then regardless of where you're at in between the plates, the potential difference will remain 693v (as long as the field is assumed to be constant). So then the problem becomes very easy:
No. E is constant let me use words . So the potential difference across the distance of the potential different is constant. The potential difference is 693 volts and it is across a distance of 0.0785 meters so E is ...
Got It?

#### mhrob24

Yes, E = 693V/.0785m

However, this just doesn't make any sense to me. Sorry if I'm being annoying but I don't want to just be satisfied getting the correct answer without seeing how it makes sense. This figure in my textbook is what is confusing me:

So, if these were the plates in my problem, then the voltage in the equation E = V/D would be the voltage from A (lets call this the zero volt plate) and b (the plate at 10 cm away from the other plate). However, that's not the voltage I am given in the problem. I am given the difference of 693v at a distance of 7.85 cm away from the zero volt plate. So this isn't the voltage from A to B, because the distance from A to B would be 10 cm, and this voltage is for a distance of 7.85 cm......So being able to use that distance and that given voltage in this equation seems like it shouldn't be correct.

#### gleem

Sorry if I'm being annoying but I don't want to just be satisfied getting the correct answer without seeing how it makes sense.
This is the attitude we like. Keep asking questions until you understand.

This is a weird problem. But it is meant, I think, for you to realize that if the electric field is a constant then the ratio of the potential difference between any two points separated by the same distance between the plates is the same. So if you took two points separated by 2 cm for example anywhere between the plates the potential difference would be the same.

In your figure above if you draw straight lines between the plates and parallel to the plates equally spaced the potential difference between adjacent lines is the same i.e. constant.

#### kuruman

Homework Helper
Gold Member
Let me elaborate on what I posted before. Between the plates the potential function depends linearly on the distance from the reference plate. The slope of the straight line is the constant electric field. This means that anywhere between the plates we can write the potential function as a straight line:
$$V(x)=Ex+V_0$$
where $V_0$ is the value of the potential at $x=0$. Here we are told that the potential is zero at plate 1. If we choose the origin of coordinates ($x=0$) to be also at plate 1, the equation simplifies to$$V(x)=Ex$$. If the value of the potential at a given point and the point are given, then you can easily find the slope $E$. The plate separation is irrelevant.

#### mhrob24

Ok, ok. I think I've got it now:

So when calculating the E field here, we can use the given voltage of 693v in the equation even though that isn't the total voltage between the actual distance between the plates (10cm) because the electric field will remain constant between these plates.

and if I truly do understand, I should be able to answer the second part of this question, which asks me to find the total potential difference between these plates. So for that, I can use the same E-field magnitude in the equation V = E*D (since the field will be the same). So because the difference at 7.85cm away from the zero volt plate is 693v, I just need to find the voltage across a distance of 2.15 cm (the remaining distance between the plates since 7.85 + 2.15 = 10cm). So:

E = 693 / .0785 = 8828.025 v/m

V = 8828.025(.0215) = 189.802 J/C

So the total potential difference between the plates becomes: 693 +189.802 = 882.802 J/C?

#### kuruman

Homework Helper
Gold Member
You still seem a bit confused about the difference between "Potential" and "Potential difference". The two concepts are related the same way as "Position" and "Distance". In either case the first concept requires a reference point (origin) while the second concept is a difference between values at two different points.

Also, if you have an expression for the function, you don't need to calculate two differences to find its value at a third point. To be specific: you have established that E = 8830 V/m. This means that $V(x)=8830~\mathrm{(V/m)}~x$. To find the potential at the other plate, just replace x with 0.100 m in the equation. Because the first plate is specified to be at zero potential, the potential difference between the first plate and any point coincides with the value of the potential at any point, but in general this will not be the case.

#### mhrob24

Yes. Its making sense now. So JUST the potential is like saying " here is a charge at (x1) distance from the origin. This has a potential of V = E*x1.

Potential DIFFERENCE is " here is the same charge that was at (x1) distance from the origin. Now I move this charge to a distance (x2) from the origin. The potential DIFFERENCE is the difference between point x1 (where the charge was initially) and x2 (when the I moved the charge to) which is ΔV = V(x2) - V(x1)

Correct?

#### kuruman

Homework Helper
Gold Member
You don't really need charges to explain this other than the charges that set up the electric potential. Let's try a different analogy using gravity.
The gravitational potential energy (GPE) of mass $m$ at height $y$ is given by $U_g=mgy$ if we take the zero of GPE at $y=0$.
The electrical potential energy (EPE) of charge $q$ at height $y$ above the plate is given by $U_e=qEy$ if we take the zero of EPE at $y=0$.

The gravitational potential is defined as GPE per unit mass, $V_g=U_g/m=gy.$
The electric potential is defined as EPE per unit charge, $V_e=U_e/q=Ey.$

The gravitational potential difference between two points is $\Delta V_g=gy_2-gy_1=g(y_2-y_1).$
The electric potential difference between two points (also known as voltage) is $\Delta V_e=Ey_2-Ey_1=E(y_2-y_1).$

The gravitational potential energy difference of mass $m$ between two points is $\Delta U_g=m\Delta V_g=mg(y_2-y_1).$
The electric potential energy difference of charge $q$ between two points is $\Delta U_e=q\Delta V_e=qE(y_2-y_1).$

Do you see now?

#### mhrob24

Lmao I really hope so....

So basically the electric potential difference does not depend on the charge moving withing the field between the plates. It only depends on the distance between two separate points within the electric field. So basically, its the electric potential energy difference of taking a charge between the plates and moving it from one point to another, divided by that charges' magnitude.

#### kuruman

Homework Helper
Gold Member
Correct. I think you're getting it now.

"Find the E-Field between plates when given only given part of the total voltage"

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