Question

Charges of 3.00 nC, -2.00 nC, -7.00 nC, and 1.00 nC are included inside a rectangular container with length 1.00 m, width 2.00 m, and height 2.50 m. Outside the container are charges of 1.00 nC and 4.00 nC. What is the electric flux through the surface of the box?

Answer 1

Answer:

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Answer 2

Answer:

The electric flux through the surface of the box$\phi _{E} =-5.65\times 10^{-2}Nm^{2}/C$

Explanation:

Electric Flux:

The electric flux through a surface describes the number of electric field lines that cross that surface. Mathematically, it is defined as the surface integral of the electric field through that surface.

The electric flux through a closed surface equals  to-

$\phi _{E} =EA\\\phi _{E} =\frac{q}{4\pi \epsilon_{o} r^{2} } 4\pi r^{2}\\\phi _{E} =\frac{q}{\epsilon_{o}}$

​where E is electric field

A is area

q is the total charge contained within the surface.

Given Charges inside the rectangular container

3.00 nC, -2.00 nC, -7.00 nC, and 1.00 nC

Find: What is the electric flux through the surface of the box?

Total charge, $q=(3-2-7+1)nC$

$\\q=-5nC$

So, electric flux through the surface of the box

$\phi _{E} =\frac{q}{\epsilon_{o}}\\\phi _{E} =\frac{-5\times 10^{-9} C}{8.85\times10^{-32} C^{2}/Nm^{2} } \\\phi _{E} =-5.65\times 10^{-2}Nm^{2}/C$

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