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2 answers:
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Answer:
1) Ex(P) = -8.34602 N/C
2) = -5.23850174216 N/C
3) Question (3) is a similar question to (1)
4) = -5.23850174216 N/C
5) ≈ 1.041466 C/m²
6) σₐ ≈ 2.2330413 C/m²
Explanation:
The given parameter of the point charge located at the center of a conducting shell
The charge of the point charge, q₁ = -8.9 μC
The inner radius of the shell, a = 2.8 cm
The outer radius of the shell, b = 4.1 cm
The charge of the conducting shell, q₂ = 2.2 μC
Therefore, we have;
1) The point P(8.5, 0)
By plugging in the values, we have;
For
R₁ < R₂ < r, for the electric field at the point, 'P', we have;
Ex(P) = -8.34602 N/C
2) For the point given with coordinates (8.5, 0), the distance of the y-component of point from the center = 0
The y-component of the electric field = 0 N/C
4) For r = 1.4 cm, along the y-axis we have;
R₁ < r < R₂
Therefore, we have;
Substituting the values, we get;
= -5.23850174216 N/C
5) The charge density, , is given as follows;
6) Similarly, we have;