HOW TO DO THIS QUESTION The electric field components due to a charge inside the cube of side 0 - Physics - Electric Charges And Fields

Question

HOW TO DO THIS QUESTION?
The electric field components due to a charge inside the
cube of side 0 1 m are as shown:

Where
α = 500 N/C − m
E y= 0,E z=
0
Calculate (i) the flux through the cube and (ii)
the charge inside the cube

Vijay Yadav · Accepted Answer

(i) Flux through the cube $= \oint \vec{E} d \vec{s} = \oint (E_{x} \hat{i} + E_{y} \hat{j} + E_{z} \hat{k}) d \vec{s} Φ = \oint (E_{x} \hat{i} d \vec{s}) [since E_{y} = E_{z} = 0] E_{x} = α x, α = 500 N/C-m E_{x} = 500 x$ A = l*l = 0 1*0 1 =0 01 m2 $So, Φ = {(\vec{E_{x}} \vec{A})}_{a t x = 0 1} + {(\vec{E_{x}} \vec{A})}_{a t x = 0 2 m} = (E_{x} A \cos 180 °)_{x = 0 1} + (E_{x} A \cos 0 °)_{x = 0 2 m} = [500 \times 0 1 \times 0 01 (- 1)] + [500 \times (0 2) \times (0 01) (1)] = - 0 5 + 1 = 0 5 {N/C -m}^{2}$ (ii) Apply Gauss's Law, $Φ = \oint \vec{E} d \vec{s} = \frac{q}{\in_{0}} \Rightarrow 0 5 = \frac{q}{\in_{0}} \Rightarrow q = 0 5 \in_{0} = 0 5 \times 8 85 \times 10^{- 12} = 4 425 \times 10^{- 12} C$

HOW TO DO THIS QUESTION? The electric field components due to a charge inside the cube of side 0.1 m are as shown: Where α = 500 N/C − m E y= 0,E z= 0 Calculate (i) the flux through the cube and (ii) the charge inside the cube

HOW TO DO THIS QUESTION?

The electric field components due to a charge inside the cube of side 0.1 m are as shown:

Where α = 500 N/C − m

E _y= 0,E _z= 0

Calculate (i) the flux through the cube and (ii) the charge inside the cube