Icabella asked in Science & MathematicsPhysics · 2 months ago

# Calculate the net electric flux?

Part J:

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.

Part K:

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.

Part L:

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.

Thankss

Relevance
• 2 months ago

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