The electric field components in the given figure are Ex = α x1/2, Ey = Ez = 0 in which α = 800 N C-1 m-1/2. The charge within the cube if net flux through the cube is 1.05 N m2 C-1 (assume a = 0.1 m)

Question

The electric field components in the given figure are Ex = α x1/2, Ey = Ez = 0 in which α = 800 N C-1 m-1/2. The charge within the cube if net flux through the cube is 1.05 N m2 C-1 (assume a = 0.1 m) (A) 9.27 × 10-12 C (B) 9.27 × 1012 C (C) 6.97 × 10-12 C (D) 6.97 × 1012 C

Tardigrade Admin · Accepted Answer

By Gaussâs law, φ=(q/ε0) or q=φε0 =1.05×8.854×10-12 C =9.27×1012 C

Q. The electric field components in the given figure are $E_{x} = α x^{1/2}$ , $E_{y} = E_{z} = 0$ in which $α = 800 N C^{- 1} m^{- 1/2}$ . The charge within the cube if net flux through the cube is $1.05 N m^{2} C^{- 1}$ (assume $a = 0.1 m$ )

$9.27 \times 1 0^{- 12} C$

$9.27 \times 1 0^{12} C$

$6.97 \times 1 0^{- 12} C$

$6.97 \times 1 0^{12} C$

By Gauss’s law, $ϕ = \frac{q}{ε _{0}}$
or $q = ϕ ε_{0}$
$= 1.05 \times 8.854 \times 1 0^{- 12} C$
$= 9.27 \times 1 0^{12} C$

Q. The electric field components in the given figure are Ex​=αx1/2, Ey​=Ez​=0 in which α=800NC−1m−1/2. The charge within the cube if net flux through the cube is 1.05Nm2C−1 (assume a=0.1m)

9.27×10−12C

9.27×1012C

6.97×10−12C

6.97×1012C