An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When the capacito

Question

An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When the capacitor plates carry charges +Q and -Q, the capacitor stores energy U0. If the separation between the plates is doubled, how much electrical energy is stored in the capacitor?
a.4U0
b.U0/4
c.U0
d.2U0
e.U0/2

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Farah 5 years 2021-09-05T13:13:10+00:00 2 Answers 161 views 0

Answers ( )

  1. thienhuong
    0
    2021-09-05T13:14:28+00:00
    Reply

    Answer:

    d.2U0

    Explanation:

    The capacitance of a capacitor is given by the following formula

    C = \epsilon_0\frac{A}{d}

    And the electric potential energy stored in a capacitor is given by

    U = \frac{Q^2}{2C}

    Applying these formula to the given variables, the initial potential energy becomes

    U_0 = \frac{Q^2}{2}\frac{d}{\epsilon_0 A}

    If d becomes 2d, then potential energy becomes

    U_0 = \frac{Q^2}{2}\frac{2d}{\epsilon_0 A} = 2U_0

  2. Dau
    0
    2021-09-05T13:15:03+00:00
    Reply

    Answer:

    d. 2U₀

    Explanation:

    The electrical energy, U, stored in a capacitor is given by;

    U = \frac{1}{2}[tex]\frac{Q^2}{C}[/tex]            ——————(i)

    Where;

    Q = Charge on the plates of the capacitor

    C = capacitance of the capacitor = Aε₀ / d

    A = Area of the plates of the capacitor

    d = distance of separation between the plates

    ε₀ = permittivity of free space.

    Substitute C = Aε₀ / d into equation (i) as follows;

    U = \frac{1}{2} Q²d / Aε₀           ———–(ii)

    From the question;

    Electrical energy U₀, is stored when the charges on the plates are +Q and -Q.

    Substitute these into equation (ii) as follows;

    U₀ = \frac{1}{2} Q²d / Aε₀                —————-(iii)

    Now, when the distance is doubled (d = 2d), the electrical energy stored becomes U₁ which is given by;

    U₁ = \frac{1}{2} Q²(2d) / Aε₀

    U₁ = 2 x \frac{1}{2} Q²d / Aε₀          [\frac{1}{2} Q²d / Aε₀ = U₀ ————– from equation (iii)]

    U₁ = 2 x U₀

    U₁ = 2U₀

    Therefore, the electrical energy stored when the separation between the plates is doubled is twice the initial electrical energy stored.

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