A 13.0 μFμF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 VV across the plates. A piece of mat

Question

A 13.0 μFμF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 VV across the plates. A piece of material having a dielectric constant of 3.75 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?

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Nguyệt Ánh 2 days 2021-07-22T14:21:46+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-22T14:23:20+00:00

    Answer:

    a) 3.744×10^-3Joules

    b) 5.265 × 10^-2Joules

    c) 0.0489Joules

    Explanation:

    Capacitors are devices used for storing electric charge.

    Energy stored in a capacitor is expressed as Eo = 1/2CVo² where:

    C is the capacitance of the capacitor

    V is the potential difference across the capacitor

    a) Energy stored in the capacitor before the dielectric is inserted can be gotten using the formula above.

    Given C = 13×10^-6F

    Vo = 24V

    Energy stored in the capacitor = 1/2 × 13×10^-6 × 24²

    Eo = 3744×10^-6Joules

    Eo = 3.744×10^-3Joules

    b) Energy is stored in the capacitor after the dielectric is inserted can be calculated using:

    E = 1/2CV²

    New potential difference V = kVo where k is dielectric constant

    Vo is the potential difference across the capacitor with out the dielectric

    V = 3.75×24

    V = 90Volts

    E = 1/2×13×10^-6×90²

    E = 52,650×10^-6

    E = 5.265 × 10^-2Joules

    c) Energy change during insertion

    = E-Eo

    = 5.265 × 10^-2 – 3.744×10^-3

    = 0.05265-0.003744

    = 0.0489Joules

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