Calculate the magnitude of the electric field inside the solid at a distance of 9.20 cmcm from the center of the cavity. Express your answer

Question

Calculate the magnitude of the electric field inside the solid at a distance of 9.20 cmcm from the center of the cavity. Express your answer with the appropriate units.

in progress 0
Calantha 5 months 2021-09-05T09:18:45+00:00 1 Answers 26 views 0

Answers ( )

    0
    2021-09-05T09:20:43+00:00

    Answer:

    A point charge of -2.14uC  is located in the center of a spherical cavity of radius 6.55cm  inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

    a) Calculate the magnitude of the electric field inside the solid at a distance of 9.20cm  from the center of the cavity.  

    Express your answer with the appropriate units.

    ==7.3  × 10⁻⁵N/C

    Explanation:

    A point charge ,q = -2.14uC = – 2.14 × 10⁻⁶

    cavity of radius , r = 6.55cm = 6.55 × 10⁻²m

    charge density in the solid = 7.35×10⁻⁴ C/m³

    Distance from the center of the cavity,R = 9.20cm = 9.2 × 10⁻²m

    Volume of shell of charge = V = \frac{4\pi }{3} (R^3 - r^3)

    Charge on the shell ,Q = V × d

    Q = V = \frac{4\pi }{3} (R^3 - r^3)d

    Q = 4.1888 × ((9.2 × 10⁻²)³ – (6.55 × 10⁻²)³) × 7.35 × 10⁻⁴

    Q = 1.47 × 10⁻⁶N/C

    Electric field at 9.2× 10⁻²m due to shell

    E1 = kQ/R²

    = \frac{(9*10^9)(1.46 * 10^-^6)}{8.46 * 10^-^3 }

    = 1.563 × 10⁻⁶N/C

    Electric field at 9.2 × 10⁻²m due to q at center

    E2

    = \frac{(9*10^9)(2.14 * 10^-^6)}{8.46 * 10^-^3 }

    = 2.276 × 10⁻⁶N/C

    magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

    = E2- E1

    = (2.276 × 10⁻⁶) – (1.563 × 10⁻⁶)N/C

    = 0.73 × 10⁻⁶N/C

    = 7.3  × 10⁻⁵N/C

Leave an answer

Browse

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