# Equation of electric field line

1. Dec 10, 2015

### 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/C

Is it equation of electric field line?But how will it represent electric field line.Because it is only a point.

2. Dec 10, 2015

### blue_leaf77

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.

Last edited: Dec 10, 2015
3. Dec 10, 2015

### gracy

How did you know the strength?
How can we figure out that it's uniform?
How can we know the direction?

4. Dec 10, 2015

### blue_leaf77

The strength or magnitude of $\mathbf{E}$ is equal to $|\mathbf{E}|$.
It has no dependency on space coordinates.
How do you normally calculate the direction of a vector?

5. Dec 10, 2015

### gracy

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

6. Dec 10, 2015

### Staff: Mentor

No. Those are unit vectors, not coordinates.

7. Dec 10, 2015

### Vanadium 50

Staff Emeritus
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.

8. Dec 10, 2015

### Staff: Mentor

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

9. Dec 10, 2015

### gracy

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

10. Dec 10, 2015

### 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.

11. Dec 10, 2015

### gracy

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

12. Dec 10, 2015

### Staff: Mentor

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.

13. Dec 10, 2015

### 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.

14. Dec 10, 2015

### 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.

15. Dec 11, 2015

### gracy

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?

Last edited: Dec 11, 2015
16. Dec 11, 2015

### ehild

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.

Last edited: Dec 11, 2015
17. Dec 11, 2015

### gracy

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

18. Dec 11, 2015

### ehild

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?

Last edited: Dec 11, 2015
19. Dec 11, 2015

### 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 !

20. Dec 11, 2015

### gracy

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

21. Dec 11, 2015

### gracy

r is position vector.

22. Dec 11, 2015

### 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.

23. Dec 11, 2015

### ehild

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

24. Dec 11, 2015

### 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.

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25. Dec 11, 2015

### gracy

is it equivalent to the following

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