What Happens to Potential Difference When Switches Are Closed?

[22990atinesh]

Homework Statement

Two capacitors $C_1$ and $C_2$ (where $C_1 > C_2$) are charged to the same initial potential difference $ΔV_i$
, 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

The Attempt at a Solution

I'm trying to understand, How polarity of charges changed for $C_2$ when switches $S_1$ and $S_2$ are closed. How left plate of $C_2$ acquires positive charge and right plate negative charge. In which direction electrons flows, from which plate to which plate. Please explain..

 

[Vibhor]

I'm trying to understand, How polarity of charges changed for $C_2$ when switches $S_1$ and $S_2$ are closed. How left plate of $C_2$ acquires positive charge and right plate negative charge. In which direction electrons flows, from which plate to which plate. Please explain..

The bottom right plate attracts electrons from top right plate .The electrons flow from top right plate to the bottom right plate which in turn repel electrons from bottom left plate to top left plate .

 

[22990atinesh]

The bottom right plate attracts electrons from top right plate .The electrons flow from top right plate to the bottom right plate which in turn repel electrons from bottom left plate to top left plate .

I didn't understand

 

[BiGyElLoWhAt]

What Vibhor means is that the right plate of $C_1$ initially has a surplus of $e^-$, and the right plate of $C_2$ is lacking in $e^-$, if you look this up, the will talk about electron "holes" which causes the positive charge. Use the sea of electrons model of current to determine what happens.

 

[BiGyElLoWhAt]

Also, make sure to keep track of the Voltages of your system the whole time.