Solving the Two Capacitor Problem: Calculating Final Potential Difference

[Vibhor]

Homework Statement

Two capacitors $C_1$ and $C_2$ (where $C_1 > C_2$) are charged to the same initial potential difference $V_i$
, but with opposite polarity. The charged capacitors are removed from the battery, and their plates are connected as shown in Figure a. The switches $S_1$ and $S_2$ are then closed, as shown in Figure b. Find the final potential difference
$V_f$ between a and b after the switches are closed

Homework Equations

The Attempt at a Solution

Let V_f be final potential difference.

$V_f = \frac{Q_{1f}}{C_1} = \frac{Q_{2f}}{C_2}$
$Q_1-Q_2 = Q_{1f}+Q_{2f}$
$Q_1-Q_2 = Q_{1f}+\frac{C_2}{C_1}Q_{1f}$
$C_1V_i-C_2V_i = \frac{C_1+C_2}{C_1}Q_{1f}$

$V_f = \frac{C_1-C_2}{C_1+C_2}V_i$

Is this correct ?

 

[ehild]

Let V_f be final potential difference.

$V_f = \frac{Q_{1f}}{C_1} = \frac{Q_{2f}}{C_2}$
$Q_1-Q_2 = Q_{1f}+Q_{2f}$
$Q_1-Q_2 = Q_{1f}+\frac{C_2}{C_1}Q_{1f}$
$C_1V_i-C_2V_i = \frac{C_1+C_2}{C_1}Q_{1f}$

$V_f = \frac{C_1-C_2}{C_1+C_2}V_i$

Is this correct ?

Yes.

ehild

 

[Vibhor]

Many thanks ehild :)