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10 August, 05:27

A sphere of radius 5.00 cmcm carries charge 3.00 nCnC. Calculate the electric-field magnitude at a distance 4.00 cmcm from the center of the sphere and at a distance 6.00 cmcm from the center of the sphere if the sphere is a solid insulator with the charge spread uniformly throughout its volume.

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  1. 10 August, 09:16
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    a) E = 8628.23 N/C

    b) E = 7489.785 N/C

    Explanation:

    a) Given

    R = 5.00 cm = 0.05 m

    Q = 3.00 nC = 3*10⁻⁹ C

    ε₀ = 8.854*10⁻¹² C² / (N*m²)

    r = 4.00 cm = 0.04 m

    We can apply the equation

    E = Qenc / (ε₀*A) (i)

    where

    Qenc = (Vr/V) * Q

    If Vr = (4/3) * π*r³ and V = (4/3) * π*R³

    Vr/V = ((4/3) * π*r³) / ((4/3) * π*R³) = r³/R³

    then

    Qenc = (r³/R³) * Q = ((0.04 m) ³ / (0.05 m) ³) * 3*10⁻⁹ C = 1.536*10⁻⁹ C

    We get A as follows

    A = 4*π*r² = 4*π * (0.04 m) ² = 0.02 m²

    Using the equation (i)

    E = (1.536*10⁻⁹ C) / (8.854*10⁻¹² C² / (N*m²) * 0.02 m²)

    E = 8628.23 N/C

    b) We apply the equation

    E = Q / (ε₀*A) (ii)

    where

    r = 0.06 m

    A = 4*π*r² = 4*π * (0.06 m) ² = 0.045 m²

    Using the equation (ii)

    E = (3*10⁻⁹ C) / (8.854*10⁻¹² C² / (N*m²) * 0.045 m²)

    E = 7489.785 N/C
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