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The electric field components in Figure are $E_x = \alpha x^{1/2}, E_y = E_z = 0,$ in which $\alpha = 800 N/C m^{1/2}.$ Calculate the charge within the cube. Assume that $a = 0.1 m.$

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1 Answer

We can use Gauss’s law to find the total charge q inside the cube.
We have $\phi = \large\frac{q}{\in_0}$$ (or) q = \phi \in_0.$ Therefore
$q= 1.05 \times 8.854 \times 10^{-12} C$
$\quad= 9.27 \times 10^{-12}C$
Hence D is the correct answer.
answered Jun 2, 2014 by meena.p
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