Respuesta :
Answer:
a) [tex]\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}[/tex], b) [tex]\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}[/tex], c) [tex]\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}[/tex]
Explanation:
a) The net flux through the cube is:
[tex]\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}[/tex]
[tex]\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}[/tex]
b) The flux through the right face is:
[tex]\Phi_{right}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}[/tex]
[tex]\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}[/tex]
c) The flux through the left face is:
[tex]\Phi_{left}=(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}[/tex]
[tex]\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}[/tex]
The expression of the flow we can find the electric flow through the faces are:
A) Net flow fi = 0
B) Left face flow is: Ф = 31.69 [tex]\frac{N m^2}{C}[/tex]Nm ^ 2 / C
C) Right face flow is: Ф = -31.69 [tex]\frac{N m^2}{C}[/tex]
Given parameters
- The value of the electric field E = 7.5 10³ N / C
To find
a) The net flow
b) The flow of the right face
c) The flow of the left face
The flux is defined by the scalar product of the normal to the area and the electric field.
Ф = E . A
Where the bold letters indicate vectors, E is the electrioc field and A the area.
The normal of the area is a vector perpendicular to the area.
a) The net flow.
In this case, the normal of one face points to the left and the normal of the other surface points in the opposite direction, bone to the right.
Ф = E A - E A
Since the magnitude of the field is constant, this expression gives zero.
[tex]E_{net}= 0[/tex]
b) The flow on the left face.
The area of the face that is square is:
A = l²
A = 0.065²
A = 4,225 10⁻³ m²
Ф = E A
Ф = - 7.50 10³ 4.225 10⁻³
Ф = - 31.69 N / m² C
c) The flow on the right face
This face points away from the left face
Ф = + 7.5 10³ 4.225 10⁻³
Ф = 31.69 Nm² / C
In conclusion using the expression of the Electric flow we can find the flows through the faces are:
A) net flow Ф = 0
B) left face flow Ф = 31.69 Nm² / C
C) right face flow is: Фf= -31.69 Nm² / C
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