Electric field direction from high V to lower V

[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

 

[gracy]

I find one thing fishy here.
Where did that 10 cm come from?It is not given and there is no explanation about this in solution.

 

[ehild]

A distorted figure is confusing. I think, the figure shows the projections of the equipotential planes onto the (x,y) plane. These planes are parallel with the z axis, which is perpendicular to the page.
The electric field vectors are parallel with the (x,y) plane and perpendicular to the blue equipotential lines. The equipotentials are at 30°angle with the x axis, the E vectors make 90°angle with the equipotentials, so the E vectors are at 120°angles with the x axis.
The green line connecting points a and b represents the displacement in the direction of the electric field, from one equipotential to the next. Its length is the distance between the equipotential surfaces. The magnitude of the electric field is potential difference divided by the distance between the equipotentials.

 

[ehild]

I find one thing fishy here.
Where did that 10 cm come from?It is not given and there is no explanation about this in solution.

You see the marks on the x axis: They are 10 cm apart, where the equipotentials intersect the x axis.

 

[gracy]

They are 10 cm apart, where the equipotentials intersect the x axis.

But it is not given.Why 10 cm specifically not something else?

 

[ehild]

But it is not given.Why 10 cm specifically not something else?

It is given in the figure.
Why? as the problem maker wanted those distances be 10 cm.

 

[gracy]

Oh!Yes.Sorry.Sorry.Sorry.

 

[gracy]

One more thing.How can one vector (linear )be parallel to a plane?

The electric field vectors are parallel with the (x,y) plane

 

[ehild]

One more thing.How can one vector (linear )be parallel to a plane?

If the straight line does not intersect a plane then it is parallel with it.

 

[gracy]

We can see that the line(linear vector)originates from x axis.But origination can not be considered as intersection.Because intersection is like cutting the plane,right?It is my last question,promise.

 

[nasu]

What line originates from x axis?