A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A piece of materi

Question

A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A piece of material having a dielectric constant of 3.55 is placed between the plates, completely filling the space between them.Part A. How much energy is stored in the capacitor before the dielectric is inserted?
Part B. How much energy is stored in the capacitor after the dielectric is inserted?
Part C. By how much did the energy change during the insertion?

in progress 0
Đan Thu 4 years 2021-08-27T10:20:44+00:00 1 Answers 13 views 0

Answers ( )

  1. Neala
    0
    2021-08-27T10:21:44+00:00
    Reply

    a) 3.27\cdot 10^{-3} J

    b) 11.60\cdot 10^{-3} J

    c) 8.33\cdot 10^{-3} J

    Explanation:

    a)

    The energy stored in a capacitor is given by

    U=\frac{1}{2}CV^2

    where

    C is the capacitance of the capacitor

    V is the potential difference across the plates of the capacitor

    For the capacitor in this problem, before insering the dielectric, we have:

    C=13.5 \mu F = 13.5\cdot 10^{-6}F is its capacitance

    V = 22.0 V is the potential difference across it

    Therefore, the initial energy stored in the capacitor is:

    U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J

    b)

    After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

    C'=kC

    where

    k = 3.55 is the dielectric constant of the material

    C is the initial capacitance of the capacitor

    Therefore, the energy stored now in the capacitor is:

    U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2

    where:

    C=13.5\cdot 10^{-6}F is the initial capacitance

    V = 22.0 V is the potential difference across the plate

    Substituting, we find:

    U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J

    C)

    The initial energy stored in the capacitor, before the dielectric is inserted, is

    U=3.27\cdot 10^{-3} J

    The final energy stored in the capacitor, after the dielectric is inserted, is

    U'=11.60\cdot 10^{-3} J

    Therefore, the change in energy of the capacitor during the insertion is:

    \Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J

    So, the energy of the capacitor has increased by 8.33\cdot 10^{-3} J

Leave an answer

Browse

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