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Three identical capacitors are connected in series across a potential source (battery). If a charge of Q flows into this combination of capacitors, how much charge does each capacitor carry?

a.Q
b.3Q
c.Q/9
d.Q/3

Respuesta :

cjrodriguez89

Answer:

a) Q

Explanation:

  • If the capacitors are connected in series, across a potential source (battery) , the battery voltage is divided between the capacitors, as follows:

       [tex]V = V_{1} + V_{2} +V_{3}[/tex]

  • By definition, each voltage is related with the capacitance and the charge, as follows:

        [tex]V = V_{1} + V_{2} +V_{3} = \frac{Q_{1} }{C_{1}} + \frac{Q_{2} }{C_{2}} + \frac{Q_{3} }{C_{3}}[/tex]

  • Now, due to the principle of conservation of charge, this can't neither accumulate nor dissapear at any point  in the circuit, so the charge must be the same for  each capacitor:

       [tex]Q = Q_{1} = Q_{2} = Q_{3}[/tex]

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