What is the connection between electric field strength and potential gradient?

[Arshad_Physic]

Homework Statement

What is the relationship between electric field strength and the potential gradient?

Homework Equations

The Attempt at a Solution

This is my Calc based Physcis lab question but I am at a total loss. I do not understand what potential gradient is in the first place. From google I figured out that it is rate of change.

My understanding is as the gradient decreases the electric field strength increases. But I know that by looking at graph I have made. I do not understand why is it so theoritically.

Please help!

Thanks,

Arshad

 

[Chewy0087]

http://en.wikipedia.org/wiki/Potential_gradient

Check that out, thinking about it more mathematically;

You agree that the electric potential is given by v = \frac{kq1q2}{r^2} , right?

Now when looking for the potential gradient, you're looking for the change of potential per distance moved away, so if for instance you plotted a graph of electric potential against distance away, the potential gradient is basically the gradient of that graph, so \frac{dV}{dr} .

Now find \frac{dV}{dr} where V = \frac{kq1q2}{r^2}

you might see a similarity with E.

 

[Arshad_Physic]

THanks! :) This makes much more sense to me now! :)

So basically, the more steeper the gradient (or equipotential lines) is, the mroe stronger the electric field gets, right?

 

[collinsmark]

http://en.wikipedia.org/wiki/Potential_gradient

Check that out, thinking about it more mathematically;

You agree that the electric potential is given by v = \frac{kq1q2}{r^2} , right?

Now when looking for the potential gradient, you're looking for the change of potential per distance moved away, so if for instance you plotted a graph of electric potential against distance away, the potential gradient is basically the gradient of that graph, so \frac{dV}{dr} .

Now find \frac{dV}{dr} where V = \frac{kq1q2}{r^2}

you might see a similarity with E.

Actually that's not quite right.

V = - \int _P E \cdot dl,

Where P is an arbitrary path (i.e. V is the "voltage" between the two endpoints of the path),

Not that it matters much for this problem, but the potential of a point charge, with respect to infinity, is:

<br /> V = -\int _{\infty} ^{r} \frac{1}{4 \pi \epsilon _{0}} \frac{q}{r&#039; ^{2}} dr&#039;

= \frac{1}{4 \pi \epsilon _{0}} \frac{q}{r}

Note that it is a function of 1/r, not 1/r².

But none of that really matters for this problem.

I think the question is asking you to look up a formula in your book, and compare it to the gradient of the potential.

The gradient of a scalar field is specified by the \nabla operator (called the "del" operator). So the gradient of the potential is specified by \nabla V.

So look in your textbook for something that relates \nabla V to E. Although I haven't seen your textbook, I'm confident this relationship is in there.

 

[Arshad_Physic]

Well, there is one:

E = -dV/dl

OR

V = - integral [ E dl]

Thanks soo much to both of you for yours' help! :)

 

[Chewy0087]

Yeah >.< sorry Arshad, replace the 1/r^2 with a 1/r in my post, foolish on my part.

 

[Saad Ahmed]

Is anyone here to explain the Result of electric field as potential gradient
E= -dV/dr
Please Explain this result
what is affect on electric intenity as Electric potential difference increase or Decrease
And aslo what is affect on electric intenity distance r increase or Decrease
Please Help me?