Electric field direction from high V to lower V

[gracy]

http://postimg.org/image/we3bnkyqn/
I know electric field points in direction in which the potential is decreasing.
How does potential decrease in that direction?
Here is what I think
potential at point a is 10v and at point b it is 20v
and direction of E points from b to a,hence
electric field points in direction in which the potential is decreasing.

Am I thinking right?

 

[sophiecentaur]

The Field is at right angles to the Equipotential surfaces.

 

[Chandra Prayaga]

The Field is at right angles to the Equipotential surfaces.

I would add this to what sophiecentaur says:
I am quoting the first two lines of the original question:
"I know electric field points in direction in which the potential is decreasing.
How does potential decrease in that direction?"
It is not that there is some other reason for potential to decrease in that direction. The direction of the electric field and the direction of decrease of potential energy are the same. If there is an electric field pointing in a certain direction, the potential decreases in that direction. Equivalently, if the potential decreases in a particular direction, the electric field points in that direction

 

[Herman Trivilino]

I know electric field points in direction in which the potential is decreasing.
How does potential decrease in that direction?

Gravitational potential energy decreases when a ball of mass m is lowered from the ceiling to the floor. Note that it moves in the direction of the gravitational field $\vec{g}$.

Likewise, electric potential energy decreases when a particle of positive charge q moves in the direction of the electric field $\vec{E}$. Since electric potential is simply electric potential energy divided by q, it also decreases.

Now, things get scrambled up a bit when the particle has a negative charge, because then the potential energy would increase, but the electric potential would still decrease.

 

[Herman Trivilino]

Here is what I think
potential at point a is 10v and at point b it is 20v
and direction of E points from b to a, [...]
Am I thinking right?

Not quite. Imagine a particle of negligible mass and positive charge located at Point b. Release it and it wouldn't move towards Point a. It would instead move towards the 10-volt line along a path of shorter distance. The electric field $\vec{E}$ is tangent to that path at each point.

 

[gracy]

The Field is at right angles to the Equipotential surfaces.

Yes.That's what I have drawn.

 

[gracy]

The electric field E⃗ \vec{E} is tangent to that path at each point.

No,according to my book

The Field is at right angles to the Equipotential surfaces.

 

[haruspex]

Yes.That's what I have drawn.

But it doesn't look like that. If your diagram is simply 2D, your field lines are at about 60 degrees to the equipotentials. Were you attempting to show a 3D view in perspective?

 

[ehild]

http://postimg.org/image/we3bnkyqn/
I know electric field points in direction in which the potential is decreasing.

That is not enough. Imagine you are on a hill. You can go down and decrease your potential in lot of ways. You can choose a tourist way descending slowly. Or you can go straight down along the steepest direction. What direction would a water flow choose? It is analogous with the charges. They accelerate along the electric field. So what is the direction of the electric field?

How does potential decrease in that direction?
Here is what I think
potential at point a is 10v and at point b it is 20v
and direction of E points from b to a,

That is wrong.

hence
electric field points in direction in which the potential is decreasing.
View attachment 90700
Am I thinking right?

No, your logic is wrong.

 

[gracy]

Were you attempting to show a 3D view in perspective?

yes,sort of.

 

[ehild]

yes,sort of.

Then show all the three axes, x, y, z, and equipotential surfaces instead of lines.

 

[gracy]

Actually ,I don't know .In my book it was an example ("solved problem)and it has been drawn in the same manner as I posted.I wanted to know the reason.

 

[ehild]

Actually ,I don't know .In my book it was an example ("solved problem)and it has been drawn in the same manner as I posted.I wanted to know the reason.

And what was the text to the problem? Copy it fully, please.

 

[gracy]

This was the question.some equipotential surfaces are shown in the figure below What can you say about the magnitude and the direction of the electric field?http://postimg.org/image/k6cm9b9wf/

 

[ehild]

This was the question.some equipotential surfaces are shown in the figure below What can you say about the magnitude and the direction of the electric field?http://postimg.org/image/k6cm9b9wf/

Well, we can imagine that the equipotential surfaces are planes and the drawing shows the projections on the (xy) plane, so the surfaces are vertical (out of the page). There are no a, b points and E vector. You see distances along the x axis. You see the angle, that the lines enclose with the x axis. And you see the potential value along the equipotentials.
What did you learn, how is the electric field and the potential related?

 

[gracy]

There are no a, b points and E vector.

No,there are not any a and b but there is E vector present in the solution part,what I wrote in post #14 was just a question.

 

[ehild]

What is the potential difference between the equipotential surfaces shown?
Remember how the potential difference was defined with work.

 

[gracy]

The part of a solution says Electric field lines are always perpendicular to the local equipotential surface .electric field points in direction in which the potential is decreasing And then the below image was given.http://postimg.org/image/we3bnkyqn/

 

[gracy]

What is the potential difference between the equipotential surfaces shown?

10 V.

 

[gracy]

The part of a solution says Electric field lines are always perpendicular to the local equipotential surface .electric field points in direction in which the potential is decreasing.
$W$=$q$$Δv$

 

[ehild]

The part of a solution says Electric field lines are always perpendicular to the local equipotential surface .electric field points in direction in which the potential is decreasing And then the below image was given.http://postimg.org/image/we3bnkyqn/

You see, it was said that the electric field is perpendicular to the equipotential surfaces. But the figure is not clear. Have you written the whole problem? There should be some explanation of the figure.

 

[ehild]

The part of a solution says Electric field lines are always perpendicular to the local equipotential surface .electric field points in direction in which the potential is decreasing.
$W$=$q$$Δv$

I would like the definition of potential difference between two points with the work done on unit positive test charge when it moves between twose points.

 

[gracy]

But the figure is not clear.

Hope this helps
http://postimg.org/image/akdl027z3/

 

[gracy]

the definition of potential difference between two points with the work done on unit positive test charge when it moves between twose points.

When the two points are a and b.$ΔV$=$Vb$-$Va$=$W$/$q$

 

[ehild]

Hope this helps
http://postimg.org/image/akdl027z3/

No, it does not help. The best is if you write the whole solution of your book.

 

[ehild]

When the two points are a and b.$ΔV$=$Vb$-$Va$=$W$/$q$

And how do you calculate the work with force and displacement?

 

[ehild]

No, it does not help. The best is if you write the whole solution of your book.

You certainly have a camera, don't you? Take a picture of the whole problem and solution with all figures.

 

[gracy]

Question:some equipotential surfaces are shown in the figure below What can you say about the magnitude and the direction of the electric field?http://postimg.org/image/k6cm9b9wf/
solution:First,we will find the direction of the vector $E$ note that Electric field lines are always perpendicular to the local equipotential surfaces.Electric field points in direction in which the potential is decreasing .So,it will be as shown below
http://postimg.org/image/akdl027z3/
Note that we have drawn the Vector $E$ lines perpendicular to all equipotential surfaces and it is pointing in that direction in which the potential is decreasing.So,the total angle vector $E$ is making with the positive x-axis is 120 degrees as shown.
Now,how to calculate the magnitude of the electric field!For this we need to jump from one equipotential surface to another through shortest (perpendicular )distance.

Then calculate ,$E$=$\frac{change in potential}{distance traveled}$
Let us suppose we are jumping from 20V line to 10V line from $B$ to $A$.Note that the $BA$ line is the shortest length between the two equipotential surfaces.The potential difference between the two lines is (20V-10V)=10Volt.Now,calculate the length$AB$=$10Sin30$cm=$5$cm=$5$×$10^-2$ m
Therefore ,$E$=$\frac{10V}{5×10^-2}$=$20$[V/m]
Hence ,the complete answer is that the vector $E$ has the value 200V/m and makes 120 degree angle with positive x-axis.

 

[gracy]

You certainly have a camera, don't you?

You don't need to worry about all that .
I can write down the whole problem;can draw figures and then upload it.I am a hardworking girl.Just need support.

 

[nasu]

So what is your question, after all?
It seems that you have the solution. Is there anything you don't understand in it?

 

[gracy]

Now,can I ask all the doubts ?
I think "yes"
1-Question (purpose of OP).

How does potential decrease in that direction(according to solution)?

 

[Herman Trivilino]

Gracy, what are the title, author, and publisher of your book?

 

[gracy]

what are the title, author, and publisher of your book?

Why?

 

[nasu]

Well, the potential decreases in many directions.
There are two parts to the "rule" for finding the direction of the electric field:
1. It if perpendicular to the equipotential lines
2. It points the way the potential decreases, when going along the line you found at point 1.

When your perpendicular line goes from crossing the 20 V surface to the 10 V surface, the potential decreases.

The more general rule is that the electric field is related to the gradient of the potential, a vector operator which shows the direction of the steepest decrease in potential (not just any decrease).

 

[gracy]

Here is what I think
potential at point a is 10v and at point b it is 20v
and direction of E points from b to a,hence
electric field points in direction in which the potential is decreasing.

what is wrong in it?

 

[nasu]

Why do you think it should be something wrong?

 

[gracy]

Why do you think it should be something wrong?

Because ehild said so.

 

[Herman Trivilino]

Why?

Because I want to read the book. If there's an error I want to report it to the publisher so it can be corrected.

 

[nasu]

I don't think he did.
It was that initially the problem was not clear enough. He was trying to get more information.
What he said with the hill and water going down is that the potential can decrease along many paths but the electric field is in the direction of steepest decrease.
At that point it was not clear yet if those lines represent surfaces in 3D or something else.
At least this is how I understand the conversation.

 

[gracy]

Ok.So,now what you say?Am I correct?

 

[gracy]

If there's an error I

Error,Did you find any?

 

[Herman Trivilino]

Yes. In the figure you posted the electric field lines are not perpendicular to the equipotential lines. If instead it was the author's intention to show a 3-D perspective of field lines perpendicular to an equipotential surface then the lines he drew to represent that surface should all be at the same potential. Either way, it appears he screwed it up. I would need to read the book to be sure.

 

[Herman Trivilino]

This figure shows the electric field lines between a pair of parallel plates. It's a cross-section. The equipotential lines are actually the the cross-sections of equipotential surfaces. Just as the pages of a book are surfaces that are parallel to each other.

 

[gracy]

Here is what I think
potential at point a is 10v and at point b it is 20v
and direction of E points from b to a,hence
electric field points in direction in which the potential is decreasing.

Is it right?

 

[nasu]

Yes. In the figure you posted the electric field lines are not perpendicular to the equipotential lines. If instead it was the author's intention to show a 3-D perspective of field lines perpendicular to an equipotential surface then the lines he drew to represent that surface should all be at the same potential. Either way, it appears he screwed it up. I would need to read the book to be sure.

Yeah, but in the original figure they show the right angles with markers. They were missing in the OP.
Of course, these angles are not really 90 degrees as drawn but they are shown to be so. A distorted figure is not really an error.:)

 

[nasu]

Is it right?

Why do you keep asking? What point you think you may not be right?
That the potential has these values at points a and b? It is obvious from the figure that it is so.
That the potential decreases when you go from 20 V to 10 V?
Or that the electric field has the direction indicated in the figure?

You are saying the same thing as in the solution, aren't you?

 

[gracy]

Why do keep asking?

Because nobody had confirmed it .And Mr.T has raised the question of book being reliable!

 

[gracy]

Ok.Few more questions

Now,how to calculate the magnitude of the electric field!For this we need to jump from one equipotential surface to another through shortest (perpendicular )distance.

Why shortest perpendicular distance in particular?I mean is there any rule behind it?

 

[ehild]

Question:some equipotential surfaces are shown in the figure below What can you say about the magnitude and the direction of the electric field?http://postimg.org/image/k6cm9b9wf/
solution:First,we will find the direction of the vector $E$ note that Electric field lines are always perpendicular to the local equipotential surfaces.Electric field points in direction in which the potential is decreasing .So,it will be as shown below
http://postimg.org/image/akdl027z3/
Note that we have drawn the Vector $E$ lines perpendicular to all equipotential surfaces and it is pointing in that direction in which the potential is decreasing.So,the total angle vector $E$ is making with the positive x-axis is 120 degrees as shown.
Now,how to calculate the magnitude of the electric field!For this we need to jump from one equipotential surface to another through shortest (perpendicular )distance.
View attachment 90813

Gracy, these are not the original pictures. You changed them. I would like to see the photos.
The lines you labelled as E and drew arrows to them are not the E lines. The E lines make 90°angles with the equipotentials. Do you know how 90°angles look like?

 

[gracy]

he lines you labelled as E and drew arrows to them are not the E lines. The E lines make 90°angles with the equipotentials. Do you know how 90°angles look like?

these angles are not really 90 degrees as drawn but they are shown to be so. A distorted figure is not really an error.:)

It is as simple as that