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A point charge q1 = -8.9 μC is located at the center of a thick conducting shell of inner radius a = 2.8 cm and outer radius b = 4.1 cm, The conducting shell has a net charge of q2 = 2.2 μC.


1)What is Ex(P), the value of the x-component of the electric field at point P, located a distance 8.5 cm along the x-axis from q1?______N/C.

2)What is Ey(P), the value of the y-component of the electric field at point P, located a distance 8.5 cm along the x-axis from q1?__0__N/C.

3)What is Ey(P), the value of the y-component of the electric field at point P, located a distance 8.5 cm along the x-axis from q1?______N/C

4)What is Ey(R), the value of the y-component of the electric field at point R, located a distance 1.4 cm along the y-axis from q1?_______N/C

5)What is σb, the surface charge density at the outer edge of the shell?___C/m2

6)What is σa, the surface charge density at the inner edge of the shell?____C/m2

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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