# Homework Help: E field in the X direction

1. Nov 13, 2009

### WINSTEW

1. The problem statement, all variables and given/known data
The electric potential along the x-axis is V = 50x - (100/x)V, where x is in meters. What is E_x at x = 1.0m

2. Relevant equations
partial derivative of dV/dx, then plug in the value for x

3. The attempt at a solution

I understand the concept of trying to solve the equation you have to take the partial derivative of the equation with respect to x, when I do that I get 1-100(x), which is were I think I am going wrong.
Der 50x would just be 1
Der 100/x =100x^-1 = -100x

so when I put x=1 in for the equation and solve I do not get the right answer.

2. Nov 13, 2009

### Delphi51

V = 50x - (100/x)
= 50x -100*x^-1
dv/dx = 50 - 100(-1)x^-2

3. Nov 13, 2009

### WINSTEW

so when i solve for x=1 i get 150. I put that answer in and it still says its wrong but to check my signs.
I changed it to -150 and its correct... am i missing something?

4. Nov 13, 2009

### WINSTEW

nevermind its -dv/dx which would give the -150 answer thanks...

Finished that problem and now i have to do another one involving an x and a y component. Obviously you have to break the thing down into x and y but again i'm struggling on the partial derivative part.

350/(sqrt (x^2 + y^2))

5. Nov 13, 2009

### Delphi51

I missed it, too. The formula is E = - dV/dx.

6. Nov 13, 2009

### WINSTEW

so the problem is now I don't know how to partial derive the function.
350/ ( (x^2 + y^2)^-.5)

7. Nov 13, 2009

### Delphi51

Looks kind of odd - are you sure you have keyed in that expression correctly?
The way it stands, it simplifies to
f = 350*(x^2 + y^2)^0.5
To differentiate that with respect to x, you would use
u = x^2 + y^2 so f = 350*u^0.5
and the chain rule: df/dx = df/du*du/dx
In the details of that, you'll need the power rule a couple of times:
d/dx of x^n = nx^(n-1)
Give it a try!

8. Nov 13, 2009

### WINSTEW

yea the equation 350/sqrt (x^2 +y^2) is given as what voltage equals and x and y are in meters. It then asks what the strength of the electric field is at x=2 and y = 2.7. I have to partially derive the equation. Once for x and once for y. Put the values in and solve for each component. I just can't get the derivation right.

9. Nov 13, 2009

### Delphi51

First you gave 350/ ( (x^2 + y^2)^-.5)
Then you gave 350/sqrt (x^2 +y^2)
These are not the same. The square root of u is the same as u^0.5 with no minus sign.
For V = 350/sqrt (x^2 +y^2) you would use the same idea
Let u = (x^2 +y^2) so that V = 350/u^0.5 = 350*u^-0.5
dV/dx = dV/du*du/dx
Do have a go at it and show your work so we can advise you further!

10. Nov 13, 2009

### WINSTEW

ok so dV/dx = dV/du * du/dx
dV = .5*350u^-.5 = 175u^-.5
du = 2x + 2y

175u^-.5/2x+2y * 2x+2y/2

=175u^-.5/2

for the x partial derivative and
175u^-.5/2 for the y partial

doesn't look right

11. Nov 13, 2009

### Delphi51

The derivative of y = x^n is dy/dx = n*x^(n-1).
If you had y = x^3, then dy/dx would be 3*x^2.

V = 350*u^-0.5 has an "n" value of -0.5 so
dV/du = 350*(-0.5)*u^(-1.5)

When differentiating u = x² + y², partially with respect to x, you take y to be a constant. So y² is also a constant and its derivative is zero.
∂u/∂x = 2x.
It doesn't make sense to write du alone; derivatives are always a ratio of two infinitesimally small differentials like du/dx. Since we are doing partial derivatives, we really should write ∂u/∂x. You can copy symbols like that from
https://www.physicsforums.com/blog.php?b=347 [Broken].

Last edited by a moderator: May 4, 2017
12. Nov 13, 2009

### WINSTEW

that symbol thing helps;
so

dV/du = -175*u^(-1.5)
now understanding that its a ∂u/∂x with respect to x; =2x and with respect to y; =2y

but how to i complete the equation to get an equation
-175*x^(1.5)/2x and -175*y^(1.5)/2y for the x and y directions respectively?

13. Nov 13, 2009

### Delphi51

∂V/∂x = ∂V/∂u*∂u/∂x
= -175*u^(-1.5)*2x

14. Nov 13, 2009

### WINSTEW

I am lost again...

Do i just put the u value ( x^2 +y^2) back in the equation and solve for x and then do the same thing but with ∂u/∂y for y?

my x and y values were 2.0 and 2.7 respectively.

15. Nov 13, 2009

### Delphi51

u = x² + y² = 2² + 2.7² = 11.24

∂V/∂x = -175*u^(-1.5)*2x = -175*11.24^(-1.5)*2*2 = -18.5

16. Nov 13, 2009

### WINSTEW

yea i'm not getting that answer to work for the Efield at point (2,2.7)

17. Nov 13, 2009

### Delphi51

Oh, did you remember the minus sign in Ex = -∂V/∂x this time?
Also, how are you combining Ex and Ey?
"It then asks what the strength of the electric field is at x=2 and y = 2.7."
suggests the answer is a single number for the E field, not separate Ex and Ey.
If you lay out the whole calc here, I'm sure someone will check it out for you!