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?

Answers (1)

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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