Numerical - 7

Q . The electric field components in Fig. 1.27 are E(x)  = αx 1/2 , E(y) = E(z)  = 0, in which α = 800 N/C m1/2. Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a = 0.1 m.

Solution :

(a) Since the electric field has only an x component, for faces

perpendicular to x direction, the angle between E and ΔS is ± π/2. 

Therefore, the flux φ = E.ΔS is separately zero for each face

of the cube except the two shaded ones. 

Now the magnitude of the electric field at the left face is E(L)  = αx 1/2 = αa 1/2 (x = a at the left face).

The magnitude of electric field at the right face is

E(R) = α x 1/2 = α (2a) 1/2    (x = 2a at the right face).

The corresponding fluxes are

φ(L) = E(L) .ΔS = ΔS E(L) . n^ (L) =E(L) ΔS cosθ = –E(L) ΔS, 

since θ = 180°

φ(L)= –E(L) a^2

φ(R) = E(R).ΔS = E(R) ΔS cosθ = E(R) ΔS, since θ = 0°

φ(R)= E(R) a^2

Net flux through the cube = φ(R) + φ(L)  = E(R)a^2 – E(L) a^2 

= a^2 [E(R)  – E(L)] = αa^2 [(2a) ^1/2 – a ^1/2]

= αa ^5/2 [2^1/2 - 1] = 1.05 N m2 C–1

(b) We can use Gauss’s law to find the total charge q inside the cube.

We have φ = q/ε0 or q = φε 0 . Therefore,

q = 1.05 × 8.854 × 10–^12 C = 9.27 × 10^–12 C