# Find Force From A Graph Of Potential

1. Sep 8, 2007

### Winzer

1. The problem statement, all variables and given/known data
The electric potential along the x-axis (in kV) is plotted versus the value of x, (in m). Evaluate the x-component of the electrical force on a charge of 4.05 C located on the x-axis at -13.0 m.

2. Relevant equations
$$E_{x}= -\frac{\partial V}{\partial X}$$

3. The attempt at a solution

Ok So i drew a tangent line at x=-13 and kv$$\approx$$62.5

I know I must perform Eq=F for the x-axis.
So I did:
$$E_{x}=-\frac{\partial V}{\partial X}=\frac{f(x+h)-f(x)}{h}$$
$$\rightarrow\frac{f(-13+h)-f(-13)}{h}$$ I broke each cube into 5's vertically and thought 1 would be good to equall h so :
$$\frac{f(-12)-f(-13)}{1}$$ $$\rightarrow \frac{-125-62.5}{1}$$
Sound good so far? My online HW system says I have to be really accurate with these measurements and such

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2. Sep 9, 2007

### learningphysics

I'm seeing voltage as 0 at x=-12.5, and a positive value at x=-13.

3. Sep 9, 2007

### Winzer

so :
$$\frac{f(-12.5)-f(-13)}{.5}$$

Last edited: Sep 9, 2007
4. Sep 9, 2007

### Astronuc

Staff Emeritus
Correct. V(-13) > 0 and dV/dx (-12.5) < 0.

But for accuracy, one wants to minimize dx, or $\Delta x$ in this case.

Last edited: Sep 9, 2007
5. Sep 9, 2007

### learningphysics

Yeah, what you had before was ok, but h=0.5 is better than h=1... smaller the h the better...

6. Sep 9, 2007

### Winzer

$$\frac{f(-12.9)-f(-13)}{.9}= 34.7222kv/m \longrightarrow 34722.22 V/m$$

7. Sep 9, 2007

### learningphysics

Although you want h to be small, you want it to be big enough to accurately discriminate between two points... don't make h smaller than 1 division... 1 division on your graph should be good... I'm seeing 14 divisions between -15 and -10... so that's 0.357 per division.

so maybe use h = 0.357 which is 1 division horizontally.

-13 + 0.357 = -12.643

8. Sep 9, 2007

### Winzer

does f(-12.643)=125 sound accurate?

9. Sep 9, 2007

### learningphysics

f(-13) should be greater than f(-12.643)....

from your first post you measured f(-13) at 62.5...

so one of them isn't right.

Looking at your first post... you drew the tangent at x=-13 right? Did you get the slope of the line?

10. Sep 9, 2007

### Winzer

Oops. I meant -125

11. Sep 9, 2007

### learningphysics

Hmmm... that doesn't seem right to me either... don't you think f(-12.5) = 0 ? So f(-12.643) would be positive but less than f(-13)...

The best way might just be to draw the tangent at x = -13, and then get the slope of that line as you had initially mentioned...

12. Sep 9, 2007

### Winzer

But I don't think it is accurate enough

13. Sep 9, 2007

### learningphysics

what do you get as the slope?

14. Sep 9, 2007

### Winzer

from the top -1.875E2 kv/m

15. Sep 9, 2007

### learningphysics

Hmmm... that doesn't look right... where does the line intersect the y- axis. where does it intersect the x - axis?

16. Sep 9, 2007

### Winzer

ok -250Kv/m is my slope

17. Sep 9, 2007

### learningphysics

That sounds much better.

18. Sep 9, 2007

### Winzer

so I convert -250kv/m to -250000v/m times 4.05E-6 C and get -1.0125 N.
but it is wrong