Ask Question
27 July, 19:42

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 77.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 3.00*106 V/m.

Part A: What is the potential difference between the plates?

Part B:What is the area of each plate?

Part C:What is the capacitance?

+1
Answers (1)
  1. 27 July, 19:53
    0
    Part A: 7500 V

    Part B: 2.899*10⁻³ m²

    Part C: 10.27 pF or 10.27*10⁻¹² F

    Explanation:

    Part A:

    Applying,

    E = V/d ... Equation 1

    Where E = electric field intensity between the plates, V = potential difference between the plates, d = distance of separation between the plates

    make V the subject of the equation above,

    V = Ed ... Equation 2

    Given: E = 3.0*10⁶ V/m, d = 2.5 mm = 2.5*10⁻³ m

    Substitute into equation 2

    V = 3.0*10⁶ (2.5*10⁻³)

    V = 7.5*10³ V

    V = 7500 V

    Part B:

    Using,

    E = Q / (e₀A) ... Equation 3

    Where Q = Charge on each plate of the capacitor, A = Area of each plate, e₀ = constant = dielectric = permitivity of free space

    make A the subject of the equation,

    A = Q / (e₀E) ... Equation 4

    Given: Q = 77 nC = 77*10⁻⁹ C, E = 3.0*10⁶ V/m

    Constant: e₀ = 8.854*10⁻¹² F/m

    Substitute into equation 4

    A = 77*10⁻⁹ / (8.854*10⁻¹² * 3.0*10⁶)

    A = 77*10⁻⁹ / (26.562*10⁻⁶)

    A = 2.899*10⁻³ m²

    A = 2.899*10⁻³ m².

    Part C:

    Using,

    Q = CV ... Equation 5

    Where C = Capacitance of the capacitor

    make C the subject of the equation

    C = Q/V ... Equation 6

    Given: Q = 77 nC = 77*10⁻⁹ C, V = 7500 V

    Substitute into equation 6

    C = 77*10⁻⁹/7500

    C = 10.27*10⁻¹² F

    C = 10.27 pF
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 77.0 nC. The plates are in vacuum. The ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers