Question

The electrostatic potential of a certain charge distribution in Cartesian is given by Φ(x,y,z) = V₀/a³ xyz exp[-(x+y+z)/a] , where Vo and a are constants What is the electric field of this potential? Calculate the charge density of the source of this field and What is the total charge in a cube of side a with one corner at the origin and sides parallel to the axes? Write your answer as a numerical multiple of ϵo Voa

Answer 1

given

electrostatic potential

V(x,y,z) = (Vo/a3) * xyz * exp(-(x + y + z)/a)

a. E = -dV/dx i - dV/dy j - dV/dz k

E = -(Vo/a^3)[(yz*exp(-(x + y+ z)/a) - xyz*exp(-(x + y + z)/a)/a) i + (xz*exp(-(x + y+ z)/a) - xyz*exp(-(x + y + z)/a)/a)j + ((xy*exp(-(x + y+ z)/a) - xyz*exp(-(x + y + z)/a)/a)k) ]

b. now

from gauss law

for radial symemtry

E = qin/epsilon*4*pi*r^2

charge densoty = rho = 3*qin/4*pi*r^3

rho = 3(E.r)*epsilon/r^2

hence

rho(x,y,z) = -3(Vo/a^4)xyz[3a - (x + y + z)]*epsilon*exp(-(x + y+ z)/a)/(x^2 + y^2 + z^2)

c. total charge in the cube can be found as

q = integral(integral(integral(rho*dx)dy)dz) from ( 0 to (0 to (0 to x) y ) a)

answered by: anonymous