A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed uniformly th

Question

A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed uniformly throughout the volume of the shell. What is the magnitude of the electric field at a distance r = 11.2 cm from the center of the shell? (ε0 = 8.85 × 10-12 C2/N ∙ m2) (Give your answer to the nearest 0.01 MN/C)

in progress 0
Latifah 3 years 2021-09-05T13:28:55+00:00 1 Answers 368 views 0

Answers ( )

  1. Dulcie
    0
    2021-09-05T13:30:03+00:00
    Reply

    Answer:

    E = 1580594.95 N/C

    Explanation:

    To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss’ law:

    \int EdS=\frac{Q_{in}}{\epsilon_o}   (1)

    dS: differential of the Gaussian surface

    Qin: charge inside the Gaussian surface

    εo: dielectric permittivity of vacuum =  8.85 × 10-12 C2/N ∙ m2

    The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

    \int EdS=ES=E(4\pi r^2)   (2)

    Qin is calculate by using the charge density:

    Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho  (3)

    Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

    The charge density is given by:

    \rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}

    Next, you use the results of (3), (2) and (1):

    E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})

    Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

    E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}

    hence, the electric field is 1580594.95 N/C

Leave an answer

Browse

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