A capacitor is connected to a 9 V battery and acquires a charge Q. What is the charge on the capacitor if it is connected instead to an 18 V

Question

A capacitor is connected to a 9 V battery and acquires a charge Q. What is the charge on the capacitor if it is connected instead to an 18 V battery? A capacitor is connected to a 9 V battery and acquires a charge Q. What is the charge on the capacitor if it is connected instead to an 18 V battery? 2Q 4Q Q Q/2 Request Answer Part B – Potential difference in a parallel-plate capacitor

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Cherry 5 months 2021-08-09T06:59:09+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-09T07:00:10+00:00

    Answer:

    Part A:

    2Q

    Part B:

    V_a_b=\frac{Q}{C}

    Explanation:

    The capacitance is given by:

    C=\frac{Q}{V_a_b}

    Where:

    Q=Stored\hspace{3}electric\hspace{3}charge\\V_a_b=Potential\hspace{3}difference\\C=Capacitance

    So:

    Q=C V_a_b

    In the first case:

    V1_a_b=9

    And in the second case:

    V2_a_b=18

    or:

    V2_a_b=2*V1_a_b=2*9=18

    C is a constant so:

    Q_1=C*V1_a_b\\\\so\\V1_a_b=\frac{Q_1}{C} \\ \\\\and\\\\Q_2=C*V2_a_b

    Therefore:

    Q_2=C*2V1_a_b\\\\Q_2=C*\frac{2Q_1}{C} \\\\Q_2=2*Q_1

    Part B:

    To find thepotential difference in a parallel-plate capacitor just isolate V_a_b from the capacitance equation:

    V_a_b=\frac{Q}{C}

    0
    2021-08-09T07:00:11+00:00

    Answer:

    2Q

    Explanation:

    The expression for the electric charge stored in a capacitor is given as,

    Q = CV………………… Equation 1

    Where Q = Electric charge, C = Capacitance of the capacitor, V = Potential difference applied across the plates of the capacitor.

    make C the subject of the equation

    C = Q/V…………… Equation 2

    Given: Q = Q, V = 9 V

    Substitute into equation 2

    C = Q/9 F.

    If it is connected to an 18 V battery instead,

    Q = CV

    Given: C = Q/9 F, V = 18 V

    Q = (Q/9)(18)

    Q = 2Q.

    Hence the charge on the capacitor = 2Q.

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