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# Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10^3 Nm^2/C. What is the net charge inside the box?

Q 1.17 (a) Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0 \times 10^{3}\frac{Nm^{2}}{C}$. (a) What is the net charge inside the box?

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Using Gauss’s law, we know that the flux of electric field through any closed surface S is $1/\epsilon _{0}$  times the total charge enclosed by S.

i.e.   $\phi = q/\epsilon _{0}$

where, q = net charge enclosed  and  $\epsilon _{0}$ = permittivity of free space (constant)

Given, $\phi = 8.0 \times10^3\ Nm^2/C$

$\therefore q = \phi \times\epsilon _{0} = (8.0\times10^3 \times8.85\times10^{-12})\ C$

$\implies q = 70 \times10^{-9} C = \boldsymbol{0.07 \mu C}$

This is the net charge inside the box.

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