- #1

- 66

- 1

## Homework Statement

After a capacitor of capacitance ##C_1## is charged up to a voltage ##V_o##, the Battery is removed from the circuit. Another capacitor of capacitance ##C_2## is connected in series with the capaciotr ##C_1##. Find the charge on each capacitor after equilibrium is attained. (##C_1=5microF##, ##C_2=5.3microF##, ##V_o=6.50V##).

## Homework Equations

##q=CV##

## The Attempt at a Solution

I tried cracking the problem as follows:

Initially ##C_1## has a voltage of ##V_o##. After connecting ##C_2## voltages of both the capacitors will become equal. So we have ##V_1=V_2## . Now as the initial potential was ##V_o##, and as the capacitors are connected in series ##V_1+V_2=V_o##. Therefore ##V_1=V_2=V_o/2##. Hence ##q_1=C_1V/2## and ##q_2=C_2V/2##. But after substituting the values, the final figure seems wrong!

My textbook attempts this question by taking: ##V_1=V_2## and assuming that ##C_1## initially had charge ##q_o##. Therefore as ##V_1=V_2## we have ##q_1/C_1=q_2/C_2## and taking ##q_1+q_2=q_o## and simplifying they arrive at the answer.

Textbook method seems fine. But what is wrong with my method? Am I missing any concepts? (I think I am likely to be wrong at ##V_1+V_2=V_o##).