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 th

Question

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.

in progress 0
Dâu 18 mins 2021-07-22T19:53:24+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-22T19:54:32+00:00

    Answer:

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )