Question

A cube with 1 m on a side is located in the positive x-y-z octant in a Cartesian coordinate system, with one of its points located at the origin. Find the total charge contained in the cube if the charge is given by p_v = x^2 ye^-z mC/m^3

Answer 1

Answer:

4.61 mC

Explanation:

The cube has 1 m side in the positive x-y-z octant in a Cartesian coordinate system, with one of its points located at the origin. The charge density is given as:

$\rho_v=x^2ye^{-z} \ mC/m^3$

Charge density is the charge per unit length or area or volume. It is the amount of charge in a particular region.

The charge Q is given as:

$Q=\int\limits_v {\rho_v} \, dv \\Q=\int\limits_v {\rho_v} \, dv=\int\limits^2_{x=0}\int\limits^2_{y=0}\int\limits^2_{z=0} {x^2ye^{-z}} \, dxdydz$

$Q=\int\limits^2_{x=0} {x^2} \, dx \int\limits^2_{y=0} {y} \, dy \int\limits^2_{z=0} {e^{-z}} \, dz \\\\Q=(\frac{1}{3} [x^3]^2_0)(\frac{1}{2} [y^2]^2_0)(-1 [e^{-z}]^2_0)\\\\Q=\frac{-1}{6} ([x^3]^2_0)( [y^2]^2_0)( [e^{-z}]^2_0)\\\\Q=\frac{-1}{6}[2^3-0^3][2^2-0^2][e^{-2}-e^0]\\\\Q=\frac{-1}{6}(8)(4)(0.1353-1)=\frac{-1}{6}(8)(4)(-0.8647)\\\\Q=4.61\ mC$