Calculate the net electric flux?
An imaginary spherical surface of radius 5.00 cm is centered on the point x = 0, y = 0, z = 0. Calculate the net electric flux through the surface if a point charge q1 = +3.00 nC is located at x = 1.00 cm, y = 1.00 cm, z = 0.
Calculate the net electric flux through the surface if the following charges are present: the charge in part J plus a point charge q2 = -8.00 nC at x = 2.00 cm, y = 0, z = -4.00 cm.
Calculate the net electric flux through the surface if the following charges are present: the charges in part K plus a point charge q3 = +2.00 nC at x = 4.00 cm, y = -2.00 cm, z = 3.00 cm.
- nyphdinmdLv 72 months agoFavorite Answer
Gauss' Law tells you that the net flux is proportional to the net charge enclosed by a surface. Let's apply this idea.
Part J: the radial position of the charge is r = sqrt(1^2 +1^2) = sqrt(2) = 1.41 cm
Therefore the charge is contained in the sphere and by Gauss's Law
EA = q/e0 ---> A = 4*pi*R^2 where R = 5 cm = 0.05 cm thus
E = q/(4*pi*e0*R^2) = kq/R^2 where k = 9x10^9 N-m^2/C^2 so
E = 9x10^9*3*10^-9/(0,05)^2 = 1200 N/C
Part K q2 has a radial position of r1 = sqrt(2^2 +(-4)^2) = sqrt(16+4) = sqrt(20) = 4.47 cm
since r2 < R, net charge is q1 + q2 = -5nC. Thus, E can be computed simply by taking the last result and multiplying by -5/3:
E_new = E (-5/3) = -2000 N/C
Part L: Now add in in q3 at radial position r3 = sqrt(4^2 +(-2)^2) = 4.47cm < R so net charge is
qn = q1 + q2 + q3 = -3nC --> this just gives E = -1200N/C since qn = -q1
- Andrew SmithLv 72 months ago
The flux is enclosed charge/epsilon 0 in J the enclosed charge is 3.0nC, in K it is -5 nC in L the final charge is outside the sphere so it does not contribute.