Recent content by aaas

Thank you very much , it is very clear now

thanks very much :!) I got the same answer now which means that my problem was in considering the cross sectional area as a circle . But I am a little confused , If you look at this example (b) ( hollow cylinder ) we considered the cross sectional area as surface area of cylinder . what is the...

[FONT="Arial Black"]The length of a chord if you know the radius and the perpendicular distance from the chord to the circle center. This is a simple application of Pythagoras' Theorem. [tex] w= 2 \sqrt{(r-x)^2 - r^2} \\ [\tex] where r is the radius of the circle r-x is the...

Isn't the cross sectional area is the surface area of dashed line circle ?

I divide the line into infinitesimal pieces dR=\rho \frac{dL}{A} dR=\rho \frac{dx}{2\pi (R-X) d} R = 2 \int \rho \frac{dx}{2\pi (R-x) d} , integrate for x from 0 to R

Actually , this what I exactly did . but as the variable must be in the denominator the resulting integral will include In 0 which is not defined (approach to minus infinity )

Maybe ! But it is taken from serway's book . The trick here is that the current direction is perpendicular on the cylinder axis !

Homework Statement A circular disk of radius r and thickness d is made of material with resistivity p . Show that the resistance between points a and b (Fig. attached)is independent of r and is given by R=πp /2d Homework Equations R= p L / A where L and A are length and section...