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A 40-pF capacitor is charged to a potential difference of 500 V. Its terminals are then connected to those of an uncharged 10-pF capacitor. Calculate: (a) the original charge on the 40-pF capacitor; (b) the charge on each capacitor after the connection is made; and (c) the potential difference across the plates of each capacitor after the connection.

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(a) The original charge on the 40-pF capacitor is [tex]2 .0 \ \times \ 10^{-8} \ C[/tex].

(b) The charge on each capacitor after the connection is made is [tex]4 .0 \ \times \ 10^{-9} \ C[/tex].

(c)  The potential difference across the plates of each capacitor after the connection is 100 V and 400 V.

Original charge of the capacitor

The original charge on the 40-pF capacitor is calculated as follows;

[tex]Q = CV\\\\Q = 40 \times 10^{-12} \times 500\\\\Q = 2 .0 \ \times \ 10^{-8} \ C[/tex]

Charge on each capacitor

[tex]C = \frac{C_1C_2}{C_1 + C_2} \\\\C = \frac{10 \times 10^{-12} \times 40 \times 10^{-12}}{10\times 10^{-12} \ + \ 40 \times 10^{-12}} \\\\C = 8 \times 10^{-12} \ F[/tex]

[tex]Q = Q_1 = Q_2\\\\Q = 8 \times 10^{-12} \ \times \ 500\\\\Q = 4 \times 10^{-9} \ C[/tex]

Potential difference

The potential difference across the plates of each capacitor after the connection is calculated as follows;

[tex]V = \frac{Q}{C} \\\\V_1 = \frac{Q}{C_1} \\\\V_1 = \frac{4 \times 10^{-9}}{40 \times 10^{-12}} \\\\V_1 = 100 \ V\\\\V_2 = \frac{Q}{C_2} \\\\V_2 = \frac{4 \times 10^{-9}}{10 \times 10^{-12} } \\\\V_2 = 400 \ V[/tex]

Learn more about charge of a capacitor here: https://brainly.com/question/13578522

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