A spherical capacitor contains a charge of 3.40 nC when connected to a potential difference of 240.0 V. Its plates are separated by vacuum a

Question

A spherical capacitor contains a charge of 3.40 nC when connected to a potential difference of 240.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 4.10 cm.

Calculate:
a. The capacitance
b. The radius of the inner sphere.
c. The electric field just outside the surface of the inner sphere.

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Cherry 3 years 2021-08-14T19:50:13+00:00 1 Answers 17 views 0

Answers ( )

  1. phucdien
    0
    2021-08-14T19:51:47+00:00
    Reply

    Answer:

    A) 1.4167 × 10^(-11) F

    B) r_a = 0.031 m

    C) E = 3.181 × 10⁴ N/C

    Explanation:

    We are given;

    Charge;Q = 3.40 nC = 3.4 × 10^(-9) C

    Potential difference;V = 240 V

    Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m

    A) The formula for capacitance is given by;

    C = Q/V

    C = (3.4 × 10^(-9))/240

    C = 1.4167 × 10^(-11) F

    B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.

    C = (4πε_o)/(1/r_a – 1/r_b)

    Rearranging, we have;

    (1/r_a – 1/r_b) = (4πε_o)/C

    ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m

    Plugging in the relevant values, we have;

    (1/r_a – 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))

    (1/r_a) – 24.3902 = 7.8501

    1/r_a = 7.8501 + 24.3902

    1/r_a = 32.2403

    r_a = 1/32.2403

    r_a = 0.031 m

    C) Formula for Electric field just outside the surface of the inner sphere is given by;

    E = kQ/r_a²

    Where k is a constant value of 8.99 × 10^(9) Nm²/C²

    Thus;

    E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²

    E = 3.181 × 10⁴ N/C

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