I cant understand the meaning of this root

[transgalactic]

http://i42.tinypic.com/2yju26p.gif

i know that the root supposed to be a distance
but i can't see the distance from where to where??

the charges are not a single point

i don't know how they calculate this distance

??

 

[tiny-tim]

i know that the root supposed to be a distance
but i can't see the distance from where to where??

Hi transgalactic!

It's the distance from the centre of the cube …

5 + 1.25, 1.25, and 1.25

(and later 1.25² + 1.25² = 3.125)

the charges are not a single point

Yes, you're right …

but I think they're hoping it will give a good enough approximation.

 

[transgalactic]

so if its threw the center of each cube
and in the red part
they calculate from face A

we need two distances only

can you draw in the drawing what triangles they are doing for each face
how they calculate the hipotenuse ??

and why they treat every cube in the same way??

the center point of each small cube has a different distance to point p
??

 

[tiny-tim]

… and in the red part
they calculate from face A

we need two distances only

No, you find the "hypotenuse" (it's really the diagonal of a cube ) from three distances: √(x² + y² + z²) …

x = y = 1.25,

z = 5 (I assume: it's not marked) + 1.25 = 6.25

 

[transgalactic]

this is the original question:
http://i41.tinypic.com/2ytzr60.gif

ok if we calculate the center point of cube A
this is the axes
http://i44.tinypic.com/4ibl8g.gif
on the x axes its:
3.75

on y axes it:
1.25

the distan
on z axes its:
6.25

p is distant by 5 cm and i don't have any where the $$5^2$$
i can't see the triangle with 90 angle(so i could calculate its hipotenuse )


??

 

[tiny-tim]

ok if we calculate the center point of cube A
this is the axes
http://i44.tinypic.com/4ibl8g.gif
on the x axes its:
3.75

on y axes it:
1.25

the distan
on z axes its:
6.25

No, that's B (or C) … A is 1.25 1.25 and 6.25

p is distant by 5 cm and i don't have any where the $$5^2$$

that's 5cm to the centre of the surface of cube A …

to the centre of the volume of cube A, you must add another 1.25, to make 6.25

i can't see the triangle with 90 angle(so i could calculate its hipotenuse )

(btw, it's "hypotneuse", with a "y" )

it's not really a triangle, more the diagonal of a box, with sides 1.25 1.25 and 6.25

 

[transgalactic]

i can't imagine it in the given drawing,
can you draw it with axes

??

 

[transgalactic]

where is the center point in this drawing?

 

[tiny-tim]

where is the center point in this drawing?

the centre point is the centre of the cube marked A.

 

[transgalactic]

where is our (0,0,0)
point
?

 

[transgalactic]

why the multiply the denominator by 10^-2
??

 

[transgalactic]

its only one cube we have 64 of them

how they sum 64 potentials without any integral?

 

[Mark44]

its only one cube we have 64 of them

how they sum 64 potentials without any integral?

They sum the 64 potential charges by adding them up. Remember that integration is a glorified way of adding things.

 

[Mark44]

why the multiply the denominator by 10^-2
??

I believe what they are doing is changing from cm to m. 1 cm = 10^(-2) m.

 

[Mark44]

where is our (0,0,0)
point
?

It seems that both you and Tiny-tim have been using the lowest point shown on the cube (near the dimension 1.25 cm) as the origin. You can have it wherever you want it - just be consistent in using the same point for calculating distances.

 

[transgalactic]

even if i calculate the potential of one cube

how they calculate the sum of 64 potentials without making 64 equations
and without making any integral?

 

[HallsofIvy]

That has already been answered. They are treating each small cube as a "point" and summing. That gives an approximation to the integral. Do you remember how "Riemann sums" approximate the integral when you are first learning integration?

 

[transgalactic]

they say
"this part contributes ..."
"this part contributes ..." etc..

but then they do in a sum "..."i don't know Riemann approximation

i know to sum 64 potentials
or doing an integral

they didnt do neither

 

[tiny-tim]

even if i calculate the potential of one cube

how they calculate the sum of 64 potentials without making 64 equations
and without making any integral?

The original question says "Use a numerical method" …

that means a numerical approximation,

in which you use a finite sum instead of an integral, to get an approximate result.

(and yes, you do need 64 of them, but there will be a lot of quadruplicates, which will shorten the task)

 

[transgalactic]

i don't know this method

i see that they calculate the contribution only of some parts and not all 64
and then they jump to the final solution
without any explanation

 

[Mark44]

From my casual perusal of the illustration, it appears to me that the author has calculated 1/4 of the charge, and is expecting that readers are able to continue with this technique to get the rest of the charge.

 

[transgalactic]

ahhhhhhhhhhhh
now its much clearer

thanks

i wish they would mention it some where

 

[Mark44]

You are going to find that as your studies become more advanced, textbook authors and your professors are going to have higher expectations of your abilities. They will assume that you don't need as much "hand holding" as a beginning student would.