How many electrons are contained on the negatively-charged side of a capacitor when it is fullycharged if it is connected to a 1.5 V battery

Question

How many electrons are contained on the negatively-charged side of a capacitor when it is fullycharged if it is connected to a 1.5 V battery, the capacitor’s square plates are 0.25 cm x 0.25 cm, and the plates’ separation is 1 mm?

A. 1.8 x 10^9
B. 3.8 x 10^8
C. 2.5 x 10^7
D. 5.2 x 10^5
E. 7.5 x 10^4

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Yến Oanh 4 years 2021-08-08T11:16:32+00:00 2 Answers 10 views 0

Answers ( )

  1. Acacia
    0
    2021-08-08T11:17:40+00:00
    Reply

    Answer:

    Explanation:

    Capacitance of the capacitor

    C = ε₀ A / d

    (8.85 x 10⁻¹² x .25² x 10⁻⁴) / 1 x 10⁻³

    C = .553125 x 10⁻¹³ F

    Charge = capacitance x volt

    = .553125 x 10⁻¹³ x 1.5

    = .8296875 x 10⁻¹³ C

    no of electrons

    = charge / charge on one electron

    = .8296875 10⁻¹³/ 1.6 x 10⁻¹⁹

    = 5.2 x 10^5

  2. thienan
    0
    2021-08-08T11:18:23+00:00
    Reply

    Answer:

    D. 5.2 x 10^5

    Explanation:

    Given:

    voltage of the battery connected, V=1.5\ V

    sides of square capacitor, s=0.0025\m

    separation between the plates, d=10^{-3}\ m

    Now the total charge on the parallel plate capacitor is given as:

    Q=V.A\frac{\epsilon_o}{d}

    where:

    A= area of each plate

    \epsilon_0= permittivity of free space

    Q=1.5\times (0.0025^2\times 8.85\times 10^{-12}\times\frac{1}{10^{-3}} )

    Q=8.3\times 10^{-14}\ C

    Now the no. of electrons:

    n=\frac{Q}{e}

    where:

    e=1.6\times 10^{-19}\ C is the charge on an electron

    n=\frac{8.3\times 10^{-14}}{1.6\times 10^{-19}}

    n\approx5.2\times 10^{5}

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