How to find the charge on each capacitor

In summary: Q(on X)=CV( across X i.e.=3V)What would that mean for the charges on the upper plates in both capacitors? What would that mean for the total charge in the upper half of the circuit? Where would this charge come from?The capacitors are not connected in parallel. They are in series, like the cells, in the same loop.
  • #1
Fara_k2
1
0

Homework Statement


Two batteries are connected to two identical capacitors X and Y.Initially all the plates of the capacitors are uncharged.How to find charge on each Capacitor in the given diagram.Let the Capacitance of Capacitor be C.
IMG_20180131_130237.jpg

Homework Equations


1.Q=CV
2.Kirchoff's law
3.C=C1+C2 ( Equivalent Capacitance for capacitors in parallel connection)
4.E=E1+E2(Equivalent emf for cells in series)

The Attempt at a Solution


I thought that there will be no charge on Y Capacitor as it is connected to negative terminal from both sides of 0 potential.
Q(on X)=CV( across X i.e.=3V)
Therefore Q=3C. But this answer is wrong. What is wrong in my approach?
 

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  • #2
Fara_k2 said:
I thought that there will be no charge on Y Capacitor as it is connected to negative terminal from both sides of 0 potential.
Q(on X)=CV( across X i.e.=3V)
What would that mean for the charges on the upper plates in both capacitors? What would that mean for the total charge in the upper half of the circuit? Where would this charge come from?

In general you can't know the potential of a battery terminal without analyzing the circuit. All you know for sure is the difference between the terminals.
 
  • #3
Think about Conservation of charge as well as energy on the system .Then you may be able to solve this
 
  • #4
Fara_k2 said:

Homework Statement


Two batteries are connected to two identical capacitors X and Y.Initially all the plates of the capacitors are uncharged.How to find charge on each Capacitor in the given diagram.Let the Capacitance of Capacitor be C.
View attachment 219422

Homework Equations


1.Q=CV
2.Kirchoff's law
3.C=C1+C2 ( Equivalent Capacitance for capacitors in parallel connection)
4.E=E1+E2(Equivalent emf for cells in series)

The capacitors are not connected in parallel. They are in series, like the cells, in the same loop.

Fara_k2 said:

The Attempt at a Solution


I thought that there will be no charge on Y Capacitor as it is connected to negative terminal from both sides of 0 potential.
Q(on X)=CV( across X i.e.=3V)
Therefore Q=3C. But this answer is wrong. What is wrong in my approach?
The capacitor Y is connected to the negative sides of two batteries, but it does not mean that that those negative terminals are at the same zero potential.
 

Related to How to find the charge on each capacitor

1. How do you calculate the charge on a capacitor?

The charge on a capacitor can be calculated using the formula Q = CV, where Q is the charge in Coulombs, C is the capacitance in Farads, and V is the voltage across the capacitor in Volts.

2. What is the unit of charge for a capacitor?

The unit of charge for a capacitor is Coulombs (C).

3. Can the charge on a capacitor be negative?

Yes, the charge on a capacitor can be negative. This occurs when the capacitor is connected in reverse polarity, or when it is discharged below its equilibrium potential.

4. How does the charge on a capacitor change over time?

The charge on a capacitor changes over time according to the equation Q(t) = Q(0)e^(-t/RC), where Q(t) is the charge at time t, Q(0) is the initial charge, R is the resistance in Ohms, and C is the capacitance in Farads.

5. How does the charge on a capacitor affect its voltage?

The charge on a capacitor directly affects its voltage. As the charge increases, the voltage across the capacitor also increases. This relationship is described by the equation V = Q/C, where V is the voltage, Q is the charge, and C is the capacitance.

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