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bluesteels

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- 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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