 #1
bluesteels
 28
 1
 Homework Statement:

The electric field at 2 cm from the center of long copper rod of radius 1 cm has a magnitude 3 N/C and directed outward from the axis of the rod.
(b) What would be the electric flux through a cube of side 5 cm situated such that the rod passes through opposite sides of the cube perpendicularly?
 Relevant Equations:

E = 3 N/C
λ = 3.3*10^12 C/m
r = 2cm
Area of cube = 6a^2
electric flux= q/e0
Electric Flux = E*A = 5*6(0.05)^2.
when i look up at other sources they use Electric flux = q/ (8.854*10^12 [this is e]) equation
but I am confused on why the E*A equation don't work. The answer is 0.02Nm^2/C
when i look up at other sources they use Electric flux = q/ (8.854*10^12 [this is e]) equation
but I am confused on why the E*A equation don't work. The answer is 0.02Nm^2/C
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