Respuesta :
Answer:
E_{total} = -1.13 10² N / C
the sign indicates that the electric field points to the left
Explanation:
Let's start this exercise by looking for the electric field created by an infinite leaf, for this let's use Gauss's law
[tex]\Phi_E[/tex] = ∫ E. dA = [tex]q_{int}[/tex] /ε₀
Let's define a Gaussian surface that is a cylinder, the normal to the faces of the cylinder is parallel to the field created by the face inside the surface, the normal of the cylinder walls is perpendicular to the electric field so its scalar product is zero
\Phi_E = E (2A) = q_{int} /ε₀
the number 2 is due to having two faces
E = [tex]\frac{q_{int} }{A} \ \frac{1}{2 \epsilon_0 }[/tex]
the surface charge density is
σ= Q / A
we substitute
E = [tex]\frac{\sigma }{2 \epsilon_o}[/tex]
we can see that the field is independent of the distance.
Let's write the field for each leaf, remember that the field is salient for positive charges
sheet 1
E₁ = [tex]+ \frac{\sigma_1}{2 \epsilon_o}[/tex]
sheet 2
E₂ = [tex]- \frac{\sigma_2}{2 \epsilon_o}[/tex]
sheet 3
E₃ = [tex]+ \frac{\sigma_3}{2 \epsilon_o}[/tex]
at the point the Field of sheet 1 points to the right,
the field on sheet 2 points to the left and the field on sheet 3 points to the left. Tthe electric field at the midpoint is
E_ {total} = E₁ - E₂ - E₃
E_ {total} = [tex]\frac{1}{2 \epsilon_o}[/tex] (σ₁ - σ₂ -σ₃)
calculate
E_total = [tex]\frac{1}{2 \ 8.85 \ 10^{-12}}[/tex] (8.00 -4.00 -6.00) 10⁻⁹
E_total = -1.13 10² N / C
the sign indicates that the electric field points to the left
