# Mathematical method in electrical potential ?

1. Jul 14, 2013

### igraviton

Mathematical method in electrical potential ??

Hi All,
I need mathematical help from the topic electrical potential for lectures on physics by Richard Feynman.

Need some help to understand mathematical method used here.
question :
1) From electrical potential.png
( how this partial differential of ( -p/4∏ε (z/r^3)) with respect to z
⇔ -p/4∏ε (1/r^3 - 3z^2 / r^5) ?
similarly how to do partial differentiation with respect to x

2) From perpendicular_field.png
( how this E = p/4∏ε * 3z/r^5 √(x^2 + y^2) ⇔ p/4∏ε * (3 cosθsinθ/r^3)

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• ###### perpendicular_field.png
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2. Jul 15, 2013

### tiny-tim

hi igraviton!

(try using the X2 button just above the Reply box )
you use both the product rule (or the quotient rule) and the chain rule

for the chain rule, use ∂/∂z (f(r)) = ∂f(r)/dr ∂r/∂z

(and ∂r/∂z = … ? )

3. Jul 22, 2013

### igraviton

Thank you tiny-tim,
after doing partial differentiation and chain rule i got following.
d (z/r^3)/dz = 1/[r][/3] - 3z/[x^4] * dr/dz
but actual answer is 1/[r^3] - 3z^2/[x^5]

4. Jul 22, 2013

### SteamKing

Staff Emeritus
And what is the definition of 'r' in terms of x, y, and z?

5. Jul 22, 2013

### tiny-tim

(you mean ∂r/∂z)

ok … and now what is ∂r/∂z ?

6. Jul 22, 2013

### igraviton

yes Tiny-tim,

$\frac{\partial (z/r^3)}{\partial z } = \frac{1}{r^3} - \frac{3z^2}{r^5}$

tiny-tim, as you told me i did partial differentiation using quotient rule and chain rule

and i got

$\frac{\partial (z/r^3)}{\partial z } = 1/(r^3) - (3z)/r^4 \frac{\partial r}{\partial z}$

where
$r = √(x^2 + y^2 + z^2)$

Last edited: Jul 22, 2013
7. Jul 22, 2013

### tiny-tim

yes, but what is ∂r/∂z ?

(you can't just leave it there like that! )

8. Jul 22, 2013

### igraviton

I don't know !!!
what is $\frac{\partial r}{\partial z}$

I did following :

$\frac {\partial \frac {z}{r^3}}{\partial z} = \frac {1 * r^3 - z * \frac {\partial r^3}{\partial z}}{(r^3)^2}$

= $\frac {1}{r^3} - \frac {z \frac {(\partial r^3)}{\partial z}}{r^6}$

where
$\frac {\partial r^3 }{\partial z} = (\frac {\partial r^3 }{\partial z}) (\frac {\partial r}{\partial z})$

=> $3 * r^2 * \frac {\partial r}{\partial z}$

putting back I got

=> $\frac {1}{r^3} - \frac {3z}{r^4} * \frac {\partial r}{\partial z}$

9. Jul 22, 2013

### igraviton

Let me know if i am doing wrong.. thanks in advance

10. Jul 22, 2013

### tiny-tim

??
… so what is ∂r/∂z ?

11. Jul 22, 2013

### SteamKing

Staff Emeritus
Think of r as f(x,y,z) = (x^2+y^2+z^2)^(1/2)

What is ∂f/∂z?