Electric field direction from high V to lower V

[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?

 

[gracy]

What line originates from x axis?

vector E.

 

[ehild]

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.

Nothing originates from the x axis. The parallel lines are projections of the equipotential surfaces onto the (x,y) plane. They are not vectors. They extend farther, so they intersect the x axis. It is not shown as it would make the 3D figure too complicated.

 

[ehild]

vector E.

The vector E is drawn at any chosen point. If you want the electric field at any point of the x axis, you draw it there.

 

[nasu]

Oh, if you consider them to be at z=0, then they do intersect lines in the plane (x,y).
But they do not intersect the plane itself.
The vectors in the plane are a special case of "parallel" to the plane.

 

[gracy]

If you want the electric field at the origin, you draw it there.

But how would then calculation work?we should have same scenario that 30 degrees then AB etc,in order to do further calculations.

 

[ehild]

But how would then calculation work?we should have same scenario that 30 degrees then AB etc,in order to do further calculations.

What do you mean? The calculation is written quite clearly in the book.

 

[gracy]

The calculation is written quite clearly in the book.

Yes but it is for the given case but if we change the place of E,would not it hurt the calculations?

 

[ehild]

Yes but it is for the given case but if we change the place of E,would not it hurt the calculations?

No.

 

[Herman Trivilino]

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.:)

One can imagine a series of equipotential planar surfaces, spaced equally by 10 cm, where the potential difference between adjacent pairs of surfaces is 10 volts. The electric field lines in that case would all be parallel to each other, as shown in the figure. The electric field strength would be 1.0 V/cm, but the 10 cm distance would have to be measured along a line that intersects the planes at 90°. Is that what you see in the figure?

Even if what you're saying is correct it would be an excellent way to obfuscate the concept for a student of introductory physics.

You are bending over backwards to find a way to make the author's figure valid when if it's from an American high school book it's very very likely that it's filled with errors and written by authors who do not know the subject they are writing about. The books are picked by committees consisting of members who are not content experts in the subject.

Why is the OP keeping the source a secret? Gracy, if you're a high school student you don't need to keep that secret from us. I'm a former high school student myself. :)

 

[nasu]

Once the angles are marked, the situation is quite clear. The planes are not spaced by 10 cm, this is not what the figure shows.

I don't say that it is good to have such kind of figures, if this is what the book actually shows.
But it's not error but sloppiness. Again, if this is what the book actually displays.

After all, mymath teacher (yes, some of us were high school students too) used to quote Poincare (I think) with this:
"Geometry is the art of correct reasoning from incorrectly drawn figures."
But he also liked to add: "But you better draw the figure correctly"

 

[Herman Trivilino]

The planes are not spaced by 10 cm, this is not what the figure shows.

What, then, is the author's purpose in labeling that distance "10 cm"?

I would very much like to see whatever it was that the author published. Otherwise we are guessing.

 

[nasu]

10 cm is the distance between the planes intersections of two successive planes with the x axis. It's clear from the solution that he did not mean it to be distance between planes. He (author) calculates this distance as 5 cm.

 

[Herman Trivilino]

It's clear from the solution that he did not mean it to be distance between planes. He (author) calculates this distance as 5 cm.

What?! Where did you see this? I was not aware that anything more than a few different versions of the original drawing had been posted!

 

[nasu]

See post 28. Gracy wrote the solution from the book there.

 

[Dale]

Closed pending moderation.

EDIT: It will remain closed. The OP has been answered: the field lines are perpendicular to the equipotential pointing from high potential to low potential. The discussion of the quality of the figure is too jumbled to be valuable.