Question

A 7.24 ?C point charge is at the center of a cube with sides of length 0.450 m .

a) What is the electric flux through one of the six faces of the cube??

b)How would your answer to part A change if the sides were of length 0.250 m ?

c) Explain

Answer #1

here,

a)

charge enclosed ,Q = 7.24 uC

Q = 7.24 * 10^-6 C

the electric flux through one of the six faces of the cube , phi = (Q /e0) /6

phi = 7.24 * 10^-6 /( 8.85 * 10^-12 * 6) N.m^2/C

phi = 1.36 * 10^5 N.m^2/C

b)

the electric flux through one of the six faces of the cube , phi = (Q /e0) /6

phi = 7.24 * 10^-6 /( 8.85 * 10^-12 * 6) N.m^2/C

phi = 1.36 * 10^5 N.m^2/C

c)

as we know

Gauss's Law

The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.

it does not depnds on the length of sides

so, the answer remains the same

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