Equation of electric field line

[gracy]

There is a sentence

In the electric field E⃗ =(4iˆ+4jˆ) N/C, electric potential at A(4 m, 0) is more than the electric potential at B(0, 4 m)


I want to know what does it mean electric field E⃗ =(4iˆ+4jˆ) N/CIs it equation of electric field line?But how will it represent electric field line.Because it is only a point.

 

[blue_leaf77]

Because it is only a point.

It's not a point, it's a vector. $\mathbf{E} = 4\hat{i}+4\hat{j}$ physically means that there is a uniform electric field in the x-y plane directed 45 degree from the x (or y) axis with the strength $4\sqrt{2}$ N/C.

 

[gracy]

axis with the strength 4√2.

How did you know the strength?

that there is a uniform electric field

How can we figure out that it's uniform?

x-y plane directed 45 degree from the x (or y) axis

How can we know the direction?

 

[blue_leaf77]

How did you know the strength?

The strength or magnitude of $\mathbf{E}$ is equal to $|\mathbf{E}|$.

How can we figure out that it's uniform?

It has no dependency on space coordinates.

How can we know the direction?

How do you normally calculate the direction of a vector?

 

[gracy]

It has no dependency on space coordinates.

But it has i and j ,it means it depends on coordinates

 

[Dale]

But it has i and j ,it means it depends on coordinates

No. Those are unit vectors, not coordinates.

 

[Vanadium 50]

Gracy, I don't think PhysicsForums is helping you. You've asked us about vectors in the past, and now it's clear you haven't learned them.

The problem is that you immediately jump to asking a question here without having put much work into it, and when you are guided by someone towards the answer, you don't try and work it out for yourself, but instead ask for another hint. And another. And another. Eventually, you have been hinted all the way to the answer. Well, you've gotten the answer, but you haven't really learned.

You're going to have to decide if you want to learn or not. If you want to learn, you are going to have to spend more time thinking and working on your own. In this case, your starting point for basic vectors shouldn't be asking for more hints - it should be going back to your past materials and see if you can solve this using what you have already been told.

 

[jtbell]

How can we figure out that it's uniform?

It has no dependency on space coordinates.

More explicitly, the equation \vec E = 4 \hat i + 4 \hat j does not contain x and/or y.

 

[gracy]

But i is for x and j is for y,right?

 

[blue_leaf77]

An example of non-uniform vector field may look like $\mathbf{E}(x,y) = 3xy\hat{i} + y^2\hat{j}$. Now you see that both the magnitude and direction of this E field is dependent on (x,y). I hope this helps.

 

[gracy]

you mean when x an y are there in the equation of E,it is non uniform?

 

[jtbell]

But i is for x and j is for y,right?

When we use $\hat i$ and $\hat j$ in a position vector, then they are associated with x and y, e.g. $\vec r = 3 \hat i + 4 \hat j$ means x = 3 and y = 4. More generally $\vec r = x \hat i + y \hat j$.

When we use $\hat i$ and $\hat j$ in an electric field vector, then they are associated with the vector components $E_x$ and $E_y$, e.g. $\vec E = 5 \hat i + 6 \hat j$ means $E_x = 5$ and $E_y = 6$. More generally $\vec E = E_x \hat x + E_y \hat y$.

In the example above, $E_x$ and $E_y$ do not depend on position (x and y), so this $\vec E$ is uniform.

In blue_leaf77's example $\vec E = 3xy \hat i + y^2 \hat j$, we have $E_x = 3xy$ and $E_y = y^2$, both of which depend on position, so this $\vec E$ is not uniform.

Going further with blue_leaf77's example, at the position $\vec r = 3 \hat i + 4 \hat j$, i.e. at (x,y) = (3,4), we have $\vec E = 36 \hat i + 16 \hat j$. At the position $\vec r = 5 \hat i + \hat j$, i.e. at (x,y) = (5,1), we have $\vec E = 15 \hat i + \hat j$. Etc.

 

[blue_leaf77]

It may be non-technical advice but I think you should really first try to put your own effort up to the point where you are really really in impasse in dealing with the problem you are facing with all resources in your hand. Meaning, if there is still opportunity to get the answer from passive sources like books or internet site, the last one being abundant in our current age, go for this option. I just tried typing "uniform vector field" in google and the first site to result is found in this very forum you are asking the same question, it's this one https://www.physicsforums.com/threads/what-is-a-uniform-vector-field.483218/. It has been more than 4 hours since you asked what non-uniform vector means, and imagine what you would have gotten done had you done what I just did with my browser. I believe the answer to the other 2 questions you asked in post #3 can also quickly be retrieved with the help of your browser. Prioritizing to be independent is the first key to being full-fledged in whatever field one is pursuing.

 

[gracy]

Actually I do search on browser first then come up with my specific doubts/questions(may be silly at times)but today I was really busy in some personal matter .I asked the question then closed the window then I found out I got a reply,I read it and wrote what first thought came in my mind after reading that.I have also tried to search for this topic I had written the topic only to realized that there is no internet access .

I should not have done that.I am extremely sorry.I won't do it again.If any such situation comes again I will totally detach myself from studies for some time until I am all free for my studies.I promise.

 

[gracy]

In the electric field E⃗ =(4iˆ+4jˆ) N/C, electric potential at A(4 m, 0) is more than the electric potential at B(0, 4 m)

I think this statement is wrong because potential at both the points will be same
$\vec{E}$=$\frac{dV}{dr}$

Since electric field is uniform here we can use

$\vec{E}$=$\frac{V}{r}$

$V$=$\vec{E}$×$r$

As E and r both are same for given two points A and B.

Potential should also be same for both.
Right?

 

[ehild]

Right?

NO, it is not an explanation.
What does r mean? What are the points between the potential difference is asked?
Look after how is electric potential difference defined.
What is the relation between electric potential and electric field.

 

[gracy]

What does r mean?

dv/dr is the the derivative of voltage with respect to position. It represents the magnitude of the electric field at a point. r is the position of the point

What are the points between the potential difference is asked?

Actually potential is asked not potential difference.

What is the relation between electric potential and electric field.

$\vec{E}$=$\frac{dV}{dr}$

 

[ehild]

dv/dr is the the derivative of voltage with respect to position. It represents the magnitude of the electric field at a point. r is the position of the point

Actually potential is asked not potential difference.$\vec{E}$=$\frac{dV}{dr}$

How do you differentiate a function with respect to position?
The question is where the potential is greater. You have to compare potentials at two different points.
What is the potential function in this case?

 

[BvU]

Dear Gracy,
Can you help me understand your $\vec E = {dV\over dr}$ ? The thing on the left is a vector. How do I have to interpret the thing on the right ? Because of the equals sign, it has to be a vector, but I don't see it !

 

[gracy]

$\vec E$ = $\frac{dV}{\vec dr}$

 

[gracy]

r is position vector.

 

[BvU]

Doesn't work! See Ehild post

When you look up the definition of a derivative you will find that you can't apply it this way.

 

[ehild]

r is position vector.

How can the position vectors of points A(4,0) and B(0,4) the same?

 

[BvU]

My point (and Elizabeth's point as well) was that you can't differentiate wrt a vector. So you can write $$
V_A - V_B = \int_A^B \vec E {\bf \cdot } d\vec s
$$ (Note that now you have a scalar on the left side and on the right side of the = sign).

But the other way around you can only write $$
E_x = -{dV\over dx}, \ E_y = -{dV\over dy}, \ E_z = -{dV\over dz}
$$ which can also be written as $$\vec E = - \nabla V$$

In your case you have the components of $\vec E$ readily available: $E_x = 4$ and $E_y = 4$ so it's not all that difficult to find an expression for V -- except that you can't take V = 0 at infinity as a reference. In such a case, the origin can be a good place to set V = 0 there.

Back to post #1. What does "the electric field $\vec E = (4{\bf \hat\imath} + 4 {\bf \hat\jmath})$ N/C" mean ?

It means that at every point in the plane, the electric force per coulomb is a vector with an x-component of 4 and a y component of 4 . So it must point in a direction with an angle of $\pi\over 4$ wrt the positive x-axis. Hence the magnitude $4\sqrt 2$ in an earlier post.

 

[gracy]

is it equivalent to the following

 

[BvU]

Equivalent is ambiguous here.

There are two vectors in this exercise: one vector is the Eleftric field (the force as a function of position -- although here the function is a constant), the other is the position vector in the x-y plane (like the position vector of A is $(4\hat\imath+0\hat\jmath)$ .Anyway, now comes the hammer: when I move from point $A = (4\hat\imath+0\hat\jmath)$ to point $B=(0\hat\imath+4\hat\jmath)$ via a straight line, then it seems to me that $$
V_A - V_B = \int_A^B \vec E {\bf \cdot } d\vec s = 0 $$ because everywhere the $\vec E {\bf \cdot } d\vec s = 0$ !

($d\vec s = (\hat\imath - \hat\jmath)\, ds$)

--

 

[ehild]

There are position vectors, pointing to a point from the origin, and there are free vectors, representing vector quantities which are functions of position. The independent variable is position vector and the dependent variable is a free vector.

The electric field is free vector, $\vec E$ in every point P. It is the same vector at point A (position vector $\vec a$) and also at point B (position vector $\vec b$) .
The potential difference VB-VA is equal to the negative work done by the electric field when a unit positive charge moves from A to B. When the field is uniform, this work is equal to the electric field times displacement, $W=\vec E\cdot\Delta \vec r$. $\Delta r = \vec b-\vec a$. Calculate $\vec \Delta r$ and multiply it by $\vec E$.
You certainly remember what you learned about equipotential surfaces, that they are perpendicular to the field lines. Here the field lines are straight lines with slope 1. The equipotentials in the plane are lines perpendicular to the former ones, having slope -1. The displacement from $\vec a - \vec b$ is along an equipotential.

 

[gracy]

I think
$\Delta \vec r$=$\vec b$ - $\vec a$
I think there's a period "." in the text being confused as a mathematical operator.
It's two separate equations
$W$=$\vec E\cdot\Delta \vec r$

$\Delta \vec r$=$\vec b-\vec a$

Right?

 

[ehild]

I think
$\Delta \vec r$=$\vec b$ - $\vec a$
I think there's a period "." in the text being confused as a mathematical operator.

Gracy, it is a full stop at the end of the sentence.

 

[gracy]

I had written

$\vec{E}$=$\frac{dV}{dr}$

It is wrong because on left side there is a vector quantity($\vec{E}$)and on right hand side both quantities are scalar at least one quantity on right hand side should be vector.
And we know $\vec{dr}$ is position vector

Hence $\vec{E}$=$\frac{dV}{\vec{dr}}$

But it is also wrong because division of vector is not allowed as direction can not be divided.

Therefore

$dV$=$\int_A^B \vec E {\bf \cdot } d\vec r$

$V_A$- $V_B$ =$\int_A^B \vec E {\bf \cdot } d\vec r$

$\vec{E}$ is uniform and hence it can be taken out of integral

=$\vec{E}$. $\int_A^Bd\vec r$

=$\vec{E}$. $\vec b$-$\vec a$

on Putting values we get..

$V_A$- $V_B$ = $(4\hat\imath+4\hat\jmath)$.[$(4\hat\imath+0\hat\jmath)$-($0\hat\imath+4\hat\jmath)]$$V_A$- $V_B$ = $(4\hat\imath+4\hat\jmath)$.$(4\hat\imath-4\hat\jmath)$

$V_A$- $V_B$ =(4)(4)+(4)(-4)$V_A$- $V_B$=16-16=0

Potential difference between points A and B is 0.Hence the statement

In the electric field E⃗ =(4iˆ+4jˆ) N/C, electric potential at A(4 m, 0) is more than the electric potential at B(0, 4 m)

is wrong.

Right?

 

[BvU]

Right.

Oh oh Gracy, you really have a knack for threads that stretch seemingly forever

Take this from #17:

dv/dr is the the derivative of voltage with respect to position. It represents the magnitude of the electric field at a point. r is the position of the point

I think it is clear to you by now that the derivative wrt a vector has components. Because from a point $\vec r$ you can go in different directions $d\vec r$. Dividing by a vector (and $d\vec r$ is a vector) isn't defined. I know you dislike calculus, but we really can't do without. So browse (or better: study carefully) here and here . And when doing an exercise, make a full stop at a points where you can't -- at least in principle -- distinguish what a derivative or an integral in the expression entails. I grant you it's sometimes difficult to imagine (I, for one, still have that with curl, in spite of all the explanatory examples).

I seem to sense that you have less inhibition with the alternative form: If the force is $q\vec E$, the work needed to move a charge over a differential $d\vec r$ is $dW = - q\vec E \cdot d\vec r$ . In terms of potential $dV = - \vec E \cdot d\vec r$

This is in fact still a differential form (i first called this the integral form, but that is after you add the $\int$ left and right).​

Then:

Actually potential is asked not potential difference.

If the sentence says "...V_A is more than V_B ... " then that really means you are supposed to say something about the potential difference.

And:
Gradually the notion "there is a sentence" in your problem statement evolves into something like "Is the following statement correct:

In the electric field E⃗ =(4iˆ+4jˆ) N/C, electric potential at A(4 m, 0) is more than the electric potential at B(0, 4 m) "​

So why didn't you render the problem statement a bit more faithfully when starting the thread ?

Finally:
Your post #28 shows you have indeed developed mastery of the subject at hand. #27 had some (small) room for improvement: $W/q = \Delta V = -\vec E\cdot\Delta \vec r$ , with $\Delta \vec r = \vec b-\vec a$ would have avoided your confusion. There is only a slight visual difference between a \cdot that stands for the inner product of two vectors and a period (which is a bit lower on the line). (*)

But:
Since your occupation still states "pedantic student", I am glad I still have room to offer for improvement in post #28: $dV = \int_A^B \vec E {\bf \cdot } d\vec r\$ should be $\Delta V = \int_A^B dV = \int_A^B -\vec E {\bf \cdot } d\vec r\$

And if you really want it perfect you also replace $\vec E. \vec b - \vec a\$ by $\vec E\cdot \left (\vec b - \vec a\ \right )$ (so: \cdot and brackets) !



As we know: nobody's perfect...

(*) some pedantic but well meant advice:

• use right mouse button | show Math as ... | TeX commands a bit more to evolve TeX skills
• keep math equations left and right sides together inside one single $\#\#$ ... $\#\#$ block: the = should be a TeX equals sign $=$ , not a text equals sign.. Same for plus (+ vs $+$) and minus (- vs $-$).

--

 

[blue_leaf77]

you really have a knack for threads that stretch seemingly forever

If only PF had an award for this type of salience.

use right mouse button | show Math as ... | TeX commands a bit more to evolve TeX skills

Hey, I just realized this feature. Now I don't need to quote the entirety of a certain comment from which I want to just copy one or more equations.
By the way @gracy , are you sure you don't miss any negative signs in the expression for the E field vector? Where did you cite the sentence In the electric field E⃗ =(4iˆ+4jˆ) N/C, electric potential at A(4 m, 0) is more than the electric potential at B(0, 4 m) from? Because if it's a textbook, such a fundamental mistake is very rarely found.

 

[gracy]

Where did you cite the sentence In the electric field E⃗ =(4iˆ+4jˆ) N/C, electric potential at A(4 m, 0) is more than the electric potential at B(0, 4 m) from? Because if it's a textbook, such a fundamental mistake is very rarely found.

No,Actually it was a question itself.State whether the statement is wrong or right?

 

[gracy]

I am glad I still have room to offer for improvement

I am always open to improve myself.

 

[gracy]

except that you can't take V = 0 at infinity as a reference.

Why can't we take V = 0 at infinity as a reference in here
post #24

 

[BvU]

"I am always open to improve myself" I claim the same, but like with these vector derivatives in many dimensions, there are so many directions that I am forced to pick only one to work on, and let the others wait...

Why can't we take V = 0 at infinity as a reference in here
post #24

E.g. along the line y = x the E field is a constant vector, parallel with $d\vec s$ . So $\vec E \cdot d\vec s = |\vec E|ds$ and $\int |\vec E|ds = |\vec E|\int ds \ \$: 'unbounded'.

Other way to look at this: you can't get such an electric field configuration with finite equipment ( an infinitely wide and high flat-plate capacitor).

Same thing happens with infinitely long wires. Didn't we have a thread on that already ?