Question

The electric field component in the figure are Ex=800x^2N/C, Ey=Ez=0. calculate

(a)the flux through the cube

(b)the charge within the cube

Assume thate a=0.1m

Answer 1

(a) The flux through the cube , ∅ = EA

given, electric field along x - axis is Ex = 800 x² N/C

electric field at origin , $E_0$= 0 N/C

electric field at x = a , $E_a$ = 800a²

now, The flux through the cube , ∅ = (800a² - 0) × a²

= 800a⁴

= 800 × (0.1)⁴ [ given , a = 0.1 m]

= 8 × 10^-2 Nm²/C

(b) charge within the cube , Q = ∅$\epsilon_0$= 8 × 10^-2 Nm²/C × 8.85 × 10^-12 C²/Nm²

= 7.08 × 10^-13 C

Answer 2

Answer:

answer of (b) is 7.08*10^ -13 C