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Given that E=15ax-8az V/m at a point on a conductor surface, what is the surface charge density at that point? Assume b) Region y 2 is occupied by a conductor. If the surface charge on the conductor is -20 nC/m2, find D just outside the conductor.

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User Sally CJ
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Answer:

Explanation:

a) To find the surface charge density at the point on the conductor surface, we can use the equation: E = σ/ε. Where E is the electric field at the point, σ is the surface charge density, and ε is the permittivity of free space.

Given E = 15ax - 8az V/m, we can see that there is no electric field component in the y-direction. Therefore, the surface charge density must also be zero in the y-direction.

We can find the surface charge density in the x-direction by equating the x-components of the electric field and the surface charge density:

15a = σ/ε

Solving for σ, we get:

σ = 15aε

Substituting the value of ε (ε = ε0), we get:

σ = 15aε0

Therefore, the surface charge density at the point on the conductor surface is 15aε0 C/m2.

b) The electric displacement field D just outside the conductor is related to the surface charge density σ by the equation:

D = εE

where E is the electric field just outside the conductor.

Since the conductor is an equipotential surface, the electric field just outside the conductor is perpendicular to the surface and has a magnitude given by:

E = σ/ε0

Substituting this in the above equation, we get:

D = ε0 (σ/ε0)

D = σ

Substituting the value of σ (-20 nC/m2), we get:

D = -20 nC/m2

Therefore, the electric displacement field just outside the conductor is -20 nC/m2.

To answer your question, we need to consider the following terms:

1. Electric field E

2. Surface charge density σ

3. Permittivity of free space ε0

Given that E = 15ax - 8az V/m at a point on the conductor surface, we can find the surface charge density σ using the formula:

σ = ε0 * E_n

where E_n is the normal component of the electric field on the surface (which is -8az V/m in this case) and ε0 is the permittivity of free space (8.854 x 10^-12 F/m).

σ = (8.854 x 10^-12 F/m) * (-8 V/m)

σ = -71.032 x 10^-12 C/m²

Thus, the surface charge density at that point is -71.032 pC/m².

For part b), since the region y ≥ 2 is occupied by a conductor with surface charge -20 nC/m², we can find the electric displacement D just outside the conductor. D is related to the surface charge density σ using the equation:

D = σ

In this case, σ = -20 nC/m² = -20 x 10^-9 C/m².

So, D = -20 x 10^-9 C/m² just outside the conductor.

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User DTig
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