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:

$$V = -\int _{\infty} ^{r} \frac{1}{4 \pi \epsilon _{0}} \frac{q}{r' ^{2}} dr'$$

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

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?