Question

A cube of side 6.50 cm is placed in a uniform field E = 7.50 × 10^3 N/C with edges parallel to the field lines (field enters the right face, passes through, then exits the left face).(a) What is the net flux through the cube?(b) What is the flux through the right face?(c) What is the flux through the left face?

Answer 1

GIven Information:

Length of side of the cube = s = 6.50 cm = 0.065 m  

Electric field = E = 7.50×10³ N/C

Required Information:

Net flux through the cube = ?

Flux through right side = ?

Flux through left side = ?

Answer:

Net flux through the cube = 0

Flux through right side = -31.65 N.m²/C

Flux through left side = 31.65 N.m²/C

Explanation:

The flux through the cube is given by

Φ = EAcos(θ)

Where E is the Electric field, A is the area of a side of cube and θ is the angle between field line and A.

A = s² = 0.065² = 0.00422 m²

(a) What is the net flux through the cube?

The entering flux (right face) is taken as negative  and the exiting flux (left face) is taken as positive.

The net flux is the some of flux entering the cube and flux leaving the cube which means that there magnitude will be same but direction will be opposite so they will cancel out each other.

Net Flux = 0

(b) What is the flux through the right face?

Φ = 7.50×10³*0.00422*cos(180°)

Φ = 7.50×10³*0.00422*(-1)

Φ = -31.65 N.m²/C

(c) What is the flux through the left face?

Φ = 7.50×10³*0.00422*cos(0°)

Φ = 7.50×10³*0.00422*(1)

Φ = 31.65 N.m²/C

As you can see, as I said in part (a) that the two flux has same magnitude but opposite direction that is why net flux was zero.

Answer 2

Answer:

a) $\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}$, b) $\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}$, c) $\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}$

Explanation:

a) The net flux through the cube is:

$\Phi_{net}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}+(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}$

b) The flux through the right face is:

$\Phi_{right}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}$

c) The flux through the left face is:

$\Phi_{left}=(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}$