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Get Answers to all your Questions
Due to a charge inside the cube of side 0.1 m, the electric field components are shown Ex = ax, where a =400 N/C-m, Ey = 0, Ez = 0. Calculate the net flux through the cube.
Answers (1)
Given- $E_x = ax$
$a = 400\text{N/C·m},$
$L = 0.1\ \text{m}$
Solution- $E_x \text{ at } x = 0 \Rightarrow E = 400 \cdot 0 = 0\ \text{N/C},$
$\Phi_1 = 0$
$E_x \text{ at } x = 0.1\Rightarrow E = 400 \cdot 0.1 = 40\ \text{N/C}$
$A = L^2 = 0.1^2 = 0.01\ \text{m}^2$
$\Phi_2 = E \cdot A = 40 \cdot 0.01 = 0.4\ \text{Nm}^2/\text{C}$
$\Phi_{\text{net}} = \Phi_2 - \Phi_1 = 0.4 - 0 = {0.4\ \text{Nm}^2/\text{C}}$
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