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Sagot :
Answer:
1) Ex(P) = -8.34602 N/C
2) [tex]E_y(P)[/tex] = -5.23850174216 N/C
3) Question (3) is a similar question to (1)
4) [tex]E_y(P)[/tex] = -5.23850174216 N/C
5) [tex]\sigma _b[/tex] ≈ 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)
[tex]V = k \cdot \dfrac{q_1 + q_2 }{r^2}[/tex]
By plugging in the values, we have;
For
R₁ < R₂ < r, for the electric field at the point, 'P', we have;
[tex]E_x(P) = 9 \times 10^9 \times \dfrac{-8.9 \ \times 10^{-6} + 2.2 \ \times 10 ^{-6} }{(0.085 \ )^2} = -8.34602[/tex]
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;
[tex]E = k \cdot \left( \dfrac{q_1 }{r} + \dfrac{q_2}{R_2}\right)[/tex]
Substituting the values, we get;
[tex]E_y(P) = 9 \times 10^9 \times \left( \dfrac{-8.9 \times 10^{-6}}{0.014} + \dfrac{2.2 \times 10^{-6}}{0.041}\right) = -523850174216[/tex]
[tex]E_y(P)[/tex] = -5.23850174216 N/C
5) The charge density, [tex]\sigma _b[/tex], is given as follows;
[tex]\sigma_b = \dfrac{Q}{A} = \dfrac{2.2 \times 10^{-6} }{4\times \pi \times 0.041^2 } \approx 1.041466 \ C/m^2[/tex]
6) Similarly, we have;
[tex]\sigma_a = \dfrac{Q}{A_a} = \dfrac{2.2 \times 10^{-6} }{4\times \pi \times 0.028^2 } \approx 2.2330413 \ C/m^2[/tex]
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