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Question: The electric field components in the given figure are \(E_{x} = \alpha x^{1/2},E_{y} = E_{z} = 0\)in...

The electric field components in the given figure are Ex=αx1/2,Ey=Ez=0E_{x} = \alpha x^{1/2},E_{y} = E_{z} = 0in which α=800NC−1m−1/2\alpha = 800NC^{- 1}m^{- 1/2}. The charge within the cube is if net flux through the cube is 1.05⥂Nm2C−11.05 ⥂ Nm^{2}C^{- 1} (assume a = 0.1 m)

A

9.27×10−12C9.27 \times 10^{- 12}C

B

9.27×1012C9.27 \times 10^{12}C

C

6.97×10−12C6.97 \times 10^{- 12}C

D

6.97×1012C6.97 \times 10^{12}C

Answer

9.27×10−12C9.27 \times 10^{- 12}C

Explanation

Solution

: By Gauss’s law,

φ=qε0\varphi = \frac{q}{\varepsilon_{0}}

or q=φε0q = \varphi\varepsilon_{0}

= 1.05×8.857×10−1210^{- 12}C = 9.27 × 10−1210^{- 12}C