A parallel plate capacitor is charged up by a battery. The battery is then disconnected, but the charge remains on the plates. The plates ar

Question

A parallel plate capacitor is charged up by a battery. The battery is then disconnected, but the charge remains on the plates. The plates are then pulled apart. Explain whether each of the following quantities increases, decreases, or remains the same as the distance between the plates increases.
(a) The capacitance of the capacitor
(b) The potential difference between the plates
(c) The electric field between the plates
(d) The electric potential energy stored by the capacitor.

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Tài Đức 3 years 2021-07-24T04:54:30+00:00 1 Answers 246 views 0

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  1. Verity
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    2021-07-24T04:56:15+00:00
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    Given a capacitor that is charge by a battery

    So when the battery was disconnect, the charge remains on the capacitor

    One plate will have positive charge

    And the other will have a negative charge but of same magnitude

    Now, the plates are pulled apart, so the distance between them was extended

    a. The capacitance of the capacitor?

    The capacitance of a capacitor is given as

    C = εo•A/d

    Where C is the capacitance

    εo is permissivity, a constant

    A is are of plate

    d is the distance apart

    From the formula we notice that

    Capacitance is inversely proportional to distance, so this implies that as the distance is increase, the capacitance is reduce.

    Capacitance Decreases

    b. Potential difference?

    The potential difference is given as

    q = CV

    V = q/C

    Or V=Ed

    Where E is electric field

    So the potential difference is inversely proportional to the capacitance, so we know that the capacitance is reducing, then voltage will increase.

    Or, using the second relations

    V=Ed, voltage is directly proportional to distance, So, if the distance apart is increasing then, the voltage is increasing

    Voltage is increasing

    c. Electric field between plate.

    The electric field is given as

    E = kQ/r²

    So the electric field is inversely proportional to the distance square, so as the distance is an increase, the electric field is decreased.

    Electric field decreases

    d. Electric potential energy?

    U = ½CV², then, q = CV

    Then, U = ½qV

    The electric potential energy is directly proportional to the voltage and since the voltage is increasing due to increase in distance, then the electric potential energy increases.

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