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Charges of 3.00nC,-2.00 nC,-7.00 nC, and 1.00nC are contained inside a rectangular box with length 1.00m, width 2.00m, and height 2.50m. Outside the box are charges of 1.00nC and 4.00nC . What is the electric flux through the surface of the box? (a) 0(b) -5.64× 10² N . m²/C(c) -1.47 ×10³ N . m²/C(d) 1.47× 10³ N . m²/C(e) 5.64× 10² N . m²/C

Sagot :

The electric flux through the surface is 5.65×〖10〗^2 Nm^2/C ≈5.6×〖10〗^2  Nm^2/C

The electric flux through a surface describes the number of electric field lines that cross that surface. It is expressed as the surface integral of the electric field through that surface.

Gauss's law states that for a closed surface, the electric flux through that surface depends only on the total charge present inside that surface. Hence, for a closed surface, the electric flux can be computed without evaluating any integrals.

The charges are: q1=3.0nC, q2=−2.0nC, q3=−7.0nC, and q4=1.0nC

Gauss's Law provides that the net electric flux through a closed surface depends only on the net charge enclosed by the surface, by the equation:

ϕ=Qnet÷ϵ0

where,

Φ is the net electric flux through the closed surface.

Qnet is the net charge enclosed by the surface.

∈_0=8.85×〖10〗^(-12) C^2/Nm^2 is the permittivity of free space.

The net charge inside the box is:

Qnet = q1+q2+q3+q4

=3.0nC−2.0nC−7.0nC+1.0nC

= −5.0nC=−5.0×10−9C

The net electric flux through the box is:

Φ=Q_net/∈_0

Φ=(-5.0×〖10〗^(-9) C)/(8.85×〖10〗^(-12) C^2/Nm^2 )

Φ=5.65×〖10〗^2 Nm^2/C ≈5.6×〖10〗^2  Nm^2/C

To learn more about charges inside a cavity. Click, https://brainly.com/question/14776586.

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