How to Derive Voltage Across Capacitors in Series?

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When two capacitors, C_1 and C_2, are connected in series to a voltage source V_0, the total voltage across the capacitors is the sum of the individual voltages, expressed as V_0 = V_1 + V_2. The charge Q on each capacitor is related to its capacitance and voltage by the equation V = Q/C. The total capacitance for capacitors in series is given by the formula 1/C_total = 1/C_1 + 1/C_2. This leads to the relationship that the charge on each capacitor is equal, allowing for the derivation of individual voltages in terms of the total voltage and capacitances. The final equation for the total capacitance is C_total = (C_1*C_2)/(C_1+C_2).
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Homework Statement



A known voltage source V_0 is connected by a switch to two capacitors in series with capacitance C_1 and C_2. When we flip the switch, connecting the capacitors to the voltage source, we measure the voltage across each capacitor, V_1 and V_2 respectively.

I need to derive an equation relating V_1 and V_2 in terms of V_0, C_1 and C_2

Homework Equations





The Attempt at a Solution

 
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What is the equation that relates charge Q on a capacitor to the capacitance C and the voltage V?
 
It's V=Q/C.
 
Correct. Now do you know what the total capacitance is for two caps connected in series? If you know that, what can you say about the overall Q, since you know the V?
 
Q = CV.

In series,

1/Ctotal = 1/C1 + 1/C2 + 1/C3 Etc.
 
Answer for you

:redface: In Series connection the the total potential difference is the sum of potential difference across each capacitor..so
V_0=V_1 + V_2
q=CV
q/c_0=q/c_1 + q/c_2
so
c_0 = (c_1*c_2)/(c_1+c_2)
 
Last edited:
Answer for you Superdave

:redface: In Series connection the the total potential difference is the sum of potential difference across each capacitor..so
V_0=V_1 + V_2
q=CV
q/c_0=q_1 + q_2
so
c_0 = (c_1*c_2)/(c_1+c_2)
 
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