How Is Field Strength Equal to Potential Gradient in Electric Fields?

[gabloammar]

I was checking my syllabus to see if my coursebook was missing out on any of the syllabus statements, and it mostly isn't. But for this one statement the explanation in the book isn't good enough and I don't understand what it's trying to say.


The chapter is Electric Fields, and the statement is,

-- state that the field strength of the field at a point is numerically equal to the potential gradient at that point.

Can anybody explain this to me or point me to a good website where it's explained?
I tried looking but couldn't find anything :(

 

[aftershock]

If you have a function where the gradient of that function is a vector field... then the vector field is conservative and that function is the potential function. I may be slightly off I'm trying to recall a definition from calc 3.

The formula for electric force is the gradient of the electric potential energy formula.

The electric field is the force divided by charge, and the electric potential formula is the potential energy divided by charge. So it's kind of analogous.

Or if you just want to see the math you can take Kq/r and find the gradient wrt to r. This should be equivalent to the electric field formula.

 

[gabloammar]

There's too much math involved in what you just told me. [I'm thankful someone even did that much]

Isn't there any way you could tell me about that statement with the minimal amount of math involved [I don't take it as a subject that's why I'm requesting an easier explanation].

 

[technician]

The link between field strength and potential also occurs in gravitation, which in some ways is easier to grasp because we only ever seem (need) to deal with point sources.
In gravitation I use the sketch I have attached to 'show' that force = gradient (steepness) of the potential graph.
The Earth can be thought of as sitting at the bottom of a potential well with potential increasing along the curve as distance from the Earth increases (This is the 1/r curve)
If you see the potential curve as the sides of a valley then I hope that it is easy to picture the force down the valley has something to do with the steepness of the valley sides.
The steepness is of the form d/dr which will give the 1/r^2 equation for force.
The same analogy goes for point electric charges.
The electric field between parallel charged plates is uniform so the graph of potential against distance across the plates is a straight line... constant steepness... which means that the force on a charge between parallel plates is constant.
Hope this helps and matches up with what you know.

 

[Dadface]

I was checking my syllabus to see if my coursebook was missing out on any of the syllabus statements, and it mostly isn't. But for this one statement the explanation in the book isn't good enough and I don't understand what it's trying to say.


The chapter is Electric Fields, and the statement is,

-- state that the field strength of the field at a point is numerically equal to the potential gradient at that point.

Can anybody explain this to me or point me to a good website where it's explained?
I tried looking but couldn't find anything :(

Hi gabloammar,
firstly make sure that you know what is meant by a "uniform field" a "non uniform field" and what is meant by "potential" and by "field strength".Now consider the uniform field set up between two parallel metal plates.Let the plates be separated by 0.5m and let one plate be at 0V and the other plate at 100V.The potential difference between the plates =100V and the potential gradient =100/0.5=200V/m.This gradient is the electric field strength.
In general terms, we can write that the field strength between parallel plates is numerically given by:

E=V/x (V=Potential difference x= plate separation)

In a non uniform field the field strength is numerically given by:

E=dV/dx (dV/dx= potential gradient at point being considered)

To get a more detailed understanding I would advise you to sketch graphs to show how V varies with x for a uniform and non uniform fields.The gradient of your graphs at any point is the numerical value of E at that point.

 

[gabloammar]

This is perfect! Thanks a lot guys!

Edit: sorry if it this seems like spamming. I have no intent on that, just wanted to thank the posters!