A parallel-plate capacitor is connected to a battery of electric potential difference V. If the plate separation is decreased, do the follow

Question

A parallel-plate capacitor is connected to a battery of electric potential difference V. If the plate separation is decreased, do the following quantities increase, decrease, or remain the same: (a) the capacitor’s capacitance, (b) the potential difference across the capacitor, (c) the charge on the capacitor, (d) the energy stored by the capacitor, (e) the magnitude of the electric field between the plates, and (f ) the energy density of that electric field?

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bonexptip 5 years 2021-07-23T19:11:07+00:00 1 Answers 181 views 1

Answers ( )

  1. kiet
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    2021-07-23T19:12:57+00:00
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    Answer:

    a) increases.

    b) remains the same.

    c) increases.

    d) increases.

    e) increases.

    f) increases.

    Explanation:

    a)

    • Since the capacitance of a parallel-plate  depends only on geometric constants and the dielectric between the plates, we can use the following expression to asess the value of the capacitance:

           C = \epsilon_{0}*\frac{A}{d} (1)

           where ε₀ = permitivitty of  free space

                       A = area of one of the plates

                       d=  plate separation

    • As we can see, if the plate separation is decreased, the value of the capacitance must increase.

    b)  

    • Per definition the capacitance explains the relationship between the charge on one of the conductors, and the potential difference between them, as follows:

            C = \frac{Q}{V} (2)

    • Assuming that the capacitor remains connected to the battery when the plate separation is decreased, since the voltage can’t change (as it must hold the same voltage than previously since it’s directly connected to the battery) the potential difference between plates must remain the same.

    c)

    • From B, we know that V in (2) must remain constant. Since we know from (1) that C must increase, this means from (2) that Q must increase too.

    d)

    • The energy stored in the electric field between the plates can be expressed as follows in terms of the capacitance C and the potential difference V:

           U = \frac{1}{2}* C*V^{2} (3)

    • From (1) in a) and from b) we know that the capacitance C must increase whilst V must remain the same, so U in (3) must increase also.

    e)

    • In the capacitor the magnitude of the Electric field between the plates is constant, and is related to the potential difference between them by the following linear relationship:

           V = E*d (4)

    • Since we know that V must remain the same, if the distance d decreases, the electris field E must increase in the same ratio in order to keep the equation balanced.

    f)

    • The energy density of the electric field is defined as the energy stored between plates by unit volume, as follows:

           u = \frac{U}{v} = \frac{\frac{1}{2}* C*V^{2}}{A*d} =\frac{1}{2}* \epsilon_{0}*\frac{A*V^{2} }{A*d*d} = \frac{1}{2} *\epsilon_{0}*E^{2} (5)

    • Since it’s proportional to the square of  the electric field, and we know from e) that the magnitude of the electric field must increase, u must increase too.

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