# Relation b/w Electric Field Intensity and Potential Gradient

1. Dec 19, 2013

### kashan123999

1. The problem statement, all variables and given/known data

I dont seem to understand the proper intuition behind the electric field intensity potential difference relation? please can anyone explain it with solid intuition and maybe a good analogy...and can anyone give a short analogy about the concept of electric field intensity,i just have grasped the concept,but i lack confidence while explaining it to others....

2. Relevant equations

E = - delta v/delta r

3. The attempt at a solution

2. Dec 19, 2013

### Simon Bridge

The electric field intensity is just "electric field".
It is a description of how a charge would move in space with that field in it.
By Newton's Law: $m\vec a = q\vec E$

Any conservative vector field can also be described by the gradient of a scalar potential - the potential is just another way of saying the same thing, but without having to use vectors so much ... the potential is to the charged particle what the hills and valleys of the surface of the Earth are to a ball.

3. Dec 19, 2013

### rude man

Electric field is the force on a positive test charge divided by the value of the test charge. More precisely, it is the above if the test charge is arbitrarily small. The latter definition precludes the test charge affecting the electric field to be measured.

Analogy: force F of gravity on Earth's surface = -mg k, k = unit vector pointing up;
potential = P = mgh if h = height above surface;
force = - gradient of potential = - grad P = -d/dh (mgh) k = -mg k .

For electric field E due to a charge q,
potential = kq/r, r = distance from q;
E = - grad P = -d/dr (kq/r) r = kq/r^2 r, r = unit r vector & points away from q.

4. Dec 19, 2013

### Staff: Mentor

The electric field intensity vector $\vec{E}$ is conservative in the sense that the amount of work done in moving a test charge from one spatial location to another is independent of the path taken. This feature guarantees that a scalar field P exists such that the electric field intensity vector can be expressed as the gradient of the potential. Here's how it works. If
$$\vec{E}=∇P$$
then
$$dW=\vec{F}\centerdot\vec{dx}=q\vec{E}\centerdot\vec{dx}=q\left( \frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy+\frac{\partial P}{\partial z}dz\right)=qdP$$
where dW is the amount of work done in moving the test charge through the vector displacement $\vec{dx}$ and dP is an exact differential of the potential P. If we integrate this equation over any arbitrary path, we obtain:
$$W=\int{\vec{F}\centerdot\vec{dx}}=qΔP$$
So, the amount of work done in moving the test charge from one spatial location to another is independent of the path.

5. Dec 20, 2013

### rude man

OK, except E = - grad P.

6. Dec 20, 2013

### ehild

The intensity of the electric field or electric field strength at a point P is the force exerted to unit positive charge (situated at that point P. )

The infinitesimal work dW done by the electric field E when the unit positive charge displaces by dr is dW=Edr.

The work of the electric field when a unit positive charge moves from A to B is

$$W_{AB}=\int_A^B{\vec E \vec {dr}}$$

The static electric field is conservative, so the line integral does nt depend on the path. WAB is determined by the initial an final points.

So you can say that WAB = V(A)-V(B), where V is a function of the position, and A is the starting point and B is the end point. If the charge moves in the direction of the field the work is positive, V(A)>V(B) We usually define the change of a function as the final value - initial value. So we can write $$\Delta V = V(B)-V(A) =- \int_A^B{\vec E \vec {dr}}$$
You can choose the potential that it is zero at B. Then the function V(x,y,z) is equal to the work done by the field when the unit positive charge moves from a point (x,y,z) to the place where the potential is zero.

Consider a small displacement Δr from point P (position vector r. The potential changes by grad(V)Δr when the unit positive charge moves from r to r+Δr.

If A and B are close, the change of the potential is ΔV=gradient V Δr. The direction of the gradient points to the maximum increase of the potential. Choose dr in that direction, you can say that grad V = dV/dr, and then

$$\Delta V =(dV/dr) Δr=-\frac{d\int{\vec E \vec {dr}}}{dr} Δr$$

dV /dr is the highest directional derivative: it is the gradient. So grad V=-E.

When determining the gradient you can formally take the derivative with respect to vector r. For example, the potential around a point charge in the origin is V= kQ/r. r is the magnitude of the position vector r. $r=\sqrt{(\vec r)^2}$

$$E=- \frac{dV}{d\vec r}=\frac{1}{2}kQ \left(\sqrt{(\vec r)^2}\right )^{-3}2 \vec r= kQ\frac{\vec r}{|r|^3}$$

ehild

7. Dec 20, 2013

### Staff: Mentor

Oops. Yes. Thanks rude man.

8. Dec 20, 2013

### kashan123999

Ok thank you all :) but as i asked,I needed some sort of analogy OR an "intuitive explanation",not a mathematical ones,well still they are useful but if anyone of you could please provide me with some sort of analogy or intuition,then that adding to the mathematical explanation will do wonders for me

9. Dec 20, 2013

### Staff: Mentor

Well, gravitational potential energy is a great analogy. Both rude man and ehild referred to that in their responses. Per unit mass, the gravitational potential energy is gz. The gravitational force per unit mass is minus the gradient of this -giz. If this doesn't work for you, I don't know what else to try. Maybe someone else can think of something.

10. Dec 20, 2013

### ehild

Well, think of gravity. Near the surface of Earth, at height y, the gravitational potential energy is PE= mgy. The gravitational potential is the potential energy of unit mass: U=gy. The force of gravity is -mg. The "intensity of gravitational field " is the force acting on unit mass. It is -g, and -dU/dy =-g.

ehild

Edit: Chest beat me...

11. Dec 20, 2013

### Simon Bridge

I gave you an analogy when I said:
... we've all used gravity, in some way, as something you experience every day and have a reasonable intuition about, as an analogy for general potentials and how they relate to force fields.

We use math to tell you about it because "mathematics" is the language of physics.
You are going to have to get used to this.

12. Dec 23, 2013

### kashan123999

what is then the use of that language for me or for a layman who couldn't grasp it from the inside,that inside revelation of "o yes I just know how it works"...I mean then only cramming the formulas and rote learning remains,which is unfortunately happening in my and other 3rd world countries to say the least...I mean how just one Can "Get" it,it is not that easy At first unless you are already into proper concept of The Mathematics and how it's language works in translating the physical phenomena...well it is all my view,I am a teen-age kiddo,I know I may not be that logical on my part,or there would be too many faults with my opinion,but it is just what i have for now...Btw really thank you sir I really didn't read the comments properly,the analogy is good but not very well visualized/explained as I am not a native English speaker...

And here I would say i might sound like a person who is demanding everything in his plate of mind :D I hope this is not the case with me,as I really need a push with the fundamental concepts so that I might go further towards the mathematical ones,which I will try to understand myself hopefully...

13. Dec 23, 2013

### kashan123999

Or maybe I am a retard for the record

14. Dec 23, 2013

### Simon Bridge

the same as any language.
If you truly cannot understand the language then there will be some things that you will never get.
But I suspect that you have yet to grasp it ... keep going.

I did give you an analogy to help until you have got used to reading the mathematical statements as words and grammar in a language rather than as a bundle of symbols. The analogy will always be insufficient to your mind.

I'll try describing the analogy in more detail.

Consider a "flat" landscape - i.e. not worrying about the Earth being a sphere to make the math simpler.
The landscape will have hills and valleys and generally be bumpy.
We can describe the landscape by plotting it's height above a flat plane as a function: h(x,y).
Put a ball in that landscape and it will accelerate on the slopes and so on. Picture how a ball behaves as it rolls around such a form.

The force on the ball is proportional to the gradient of the hill it finds itself on.

This is how objects behave with respect to a potential.

For the gravitational force (close to the surface of the Earth), the height function h(x,y) and the potential functions are directly related: $V=gh(x,y)$ ... The force is the mass times the gradient of the potential ... which we write as $F=-m\nabla V(x,y)=-mg\nabla h(x,y)$ ... the upside-down triangle is pronounced "gradient of"... so the math tells us that saying "the force is proportional to the gradient of the potential" is the same as saying: "the force is proportional to the gradient of the height function" ... the height function has a gradient (a slope) where the ball is if the ball is on the side of a hill.

In general:
The object "rolls" from a high potential region to a low potential region just as the ball rolls downhill.
It's a bit trickier to think of a 3D potential "landscape" so practice with 1D and 2D landscapes first.

That help?
Merry Xmas btw.

15. Dec 24, 2013

### kashan123999

Thank you very very much sir :)