A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is the magnitude

Question

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is the magnitude of the electric field at a distance (a) r = 12 cm, and (b) r = 20 cm from the center of the shell? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)

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Phúc Điền 5 years 2021-08-06T11:52:02+00:00 1 Answers 23 views 0

Answers ( )

  1. Euphemia
    0
    2021-08-06T11:53:36+00:00
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    a) E = 0

    b) 3.38\cdot 10^6 N/C

    Explanation:

    a)

    We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

    \int EdS=\frac{q}{\epsilon_0}

    where

    E is the electric field

    q is the charge contained by the Gaussian surface

    \epsilon_0 is the vacuum permittivity

    Here we want to find the electric field at a distance of

    r = 12 cm = 0.12 m

    Here we are between the inner radius and the outer radius of the shell:

    r_1 = 10 cm\\r_2 = 15 cm

    However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

    This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

    q = 0

    Therefore, the magnitude of the electric field is also zero:

    E = 0

    b)

    Here we want to find the magnitude of the electric field at a distance of

    r = 20 cm = 0.20 m

    from the centre of the shell.

    Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

    Therefore, it is given by:

    E=\frac{Q}{4\pi \epsilon_0 r^2}

    where in this problem:

    Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

    r=20 cm = 0.20 m is the distance from the centre of the shell

    Substituting, we find:

    E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

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