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?