First slide

Redistribution of charges and concept of common potential

Question

Two capacitors A and B with capacities $C_{1} a n d C_{2}$ are charged to potential difference of $V_{1} a n d V_{2},$ respectively. The plates of the capacitors are connected as shown in the figure. The upper plate of A is positive and that of B is negative. An uncharged capacitor of capacitance C and falls on the free ends to complete circuit, then

Moderate

A

the final charge on each capacitor are same to each other
B

the final sum of charge on plates a and d is $C_{1} V_{1}$
C

the final sum of charge on plates b and g is $C_{2} V_{2} - C_{1} V_{1}$
D

both 1 and 3 are correct

Solution

Charge on positive plate of A is $C_{1} V_{1} .$ Charge on negative plate of B is $- C_{2} V_{2} .$ When d plate of capacitor C is connected with the plate of A, then the total charge of (d, a) plate system will be $C_{1} V_{1}$ (conservation of charge).
$Similarty on (c, s) plates, total charge will be - C_{2} V_{2} . The total charge on (b, g) plate system will be + C_{2} V_{2} - C_{1} V_{1} .$

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