Question #126221

Ten identical charges of amount 

7

 

μ

C

7 μC

 each are closed in a rectangular box. What is the electric flux through the box?


1
Expert's answer
2020-07-14T08:52:11-0400

According to the Gauss's law, the electric flux ΦE\Phi_E through the box is proportional to the total charge QQ enclosed within this box:


ΦE=Qε0\Phi_E = \dfrac{Q}{\varepsilon_0}

where ε0=8.85×1012F/m\varepsilon_0 = 8.85\times10^{-12}F/m.

The total charge is


Q=107μC=70×106CQ = 10\cdot 7\mu C = 70\times 10^{-6}C

Substituting the numerical values, get:


ΦE=70×1068.85×10127.91×106Vm\Phi_E = \dfrac{70\times10^{-6}}{8.85\times10^{-12}} \approx 7.91\times10^6V\cdot m

Answer. 7.91*10^6 V*m.


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