Question

help plz ..

Answer 1

the question is attached. it is 2.14

Answer 2

hi
i think this:\[Q=\int\limits_{a}^{b}\rho dv\] where a7b are range of charge distribution

Answer 3

in cylindrical coordinate we have this relation for element of volume\[dv=\rho d \rho d \phi dz \] so \[Q=\int\limits_{0}^{3\times10^{-4}}(-0.1/\rho ^{2}+10^{-8})\rho d \rho d \phi dz +\int\limits_{3\times10^{-4}}^{\infty}0\times \rho d \rho d \phi dz\]

Answer 4

hosein , thank you. What about z interval? The question does not specify any limit.

Answer 5

So first this integral is right except we need to introduce limits for the other two variables.
\( \phi \) goes from 0 to \( 2\pi \). And the question asks explicitly "charge" per meter, so let's choose \( z \) from 0 to 1. Hence the total charge is
\[ \int_0^1 \int_0^{2\pi} \int_0^a \frac{-0.1 \times 10^{-12} }{\rho^2 + 10^{-8}} \ d\rho \ d\phi \ dz \]
where \( a = 3 \times 10^{-4} \). Notice I've written the charge density as C/m^3.

Answer 6

Now this integral separates out in each variable and is equal to
\[ \int_0^1 dz \int_0^{2\pi} d\phi \int_0^a \frac{-0.1 \times 10^{-12}}{\rho^2 + 10^{-8}} d\rho \]
\[ = 1 . (2\pi) . (-0.1 \times 10^{-12})(10^4) [ \arctan(\rho/10^{-4}) ]_0^a \]
\[ = ... \]

Answer 7

For part b, figure out how much charge moves through a vertical plane per second. I.e., take the amount of charge per meter, and multiply it by the velocity of the beam.

Answer 8

b)for 1 electron we have\[J=ev\]where J is current density & have \[J=I/A\]A=pi rho^2 so equal to equation and find current

Answer 9

your wellcome friend

Answer 10

Guys I am really thankful for your help.. I spent more than 3 hours on this question but thank God you are there ...