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An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2.7 μC/m2. A thick, infinite conducting slab, also oriented perpendicular to the x-axis occupiees the region between a = 2.8 cm and b = 4.8 cm. The conducting slab has a net charge per unit area of σ2 = 88 μC/m2. What is Ex(P), the value of the x-component of the electric field at point P, located a distance 6.8 cm from the infinite sheet of charge?

Respuesta :

jorgezarate

Answer:

[tex]E_{total}=4.82*10^6N/C[/tex]

vector with direction equal to the axis X.

Explanation:

We use the Gauss Law and the superposition law in order to solve this problem.

Superposition Law: the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

[tex]E_{total}=E_1+E_2[/tex]

Thanks Gauss Law we know that the electric field of a infinite sheet with density of charge σ is:

[tex]E=\sigma/(2\epsilon_o)[/tex]

Then:

[tex]E_{total}=(\sigma_1+\sigma_2)/(2\epsilon_o)=(-2.7*10^{-6}+88*10^{-6})/(2*8.85*10^{-12})=4.82*10^6N/C[/tex]

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.

Answer:

The total electric field will be 5*10^5 N/C

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