Electric potential and fields-capacitors

In summary: The potential difference across the 5μF capacitor is clear -- it's fixed by the 9V battery. For the series capacitors, consider that they must both have the same amount of charge on them (any current that pushes a charge onto one of them must push the same charge onto the other because they are connected in series) and this will also be the same amount of charge that's stored on the equivalent capacitance of the pair. Find the net capacitance and thus the net charge ...then proceed by Q1 = C1V and Q2 = C2Vand a hint, for questions like these don't find answer in terms of variables but find the value of each variable at each
  • #1
smoics
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electric potential and fields--capacitors

Homework Statement


What is the potential difference across each capacitor in the figure?
What is the potential difference across each capacitor in the figure?



Homework Equations



Q=C/V
Parallel capacitors: add up
series capacitors: (1/c + 1/c...)^-1

The Attempt at a Solution


http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter27/chapter27.html

I tried using Q=C/V and got 45,36,54 microC. And I tried what they have on the site above (although it didn't really make a whole lot of sense as to why they were doing what they did). I understand the concept of how to add capacitors, but how to get their individual charges...? I also tried dividing the voltage by 3 because there are three capacitors...Thanks!
 

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  • #2


First find net capacitane and thus net charge given by battery
then find charge going in each parallel branch by using Q1/C1 = Q2/C2 (as both branches are parallel so V1=V2)

find Q1 and Q2 and thus use them and the formula Q=CV to find C for all the capacitors.
 
  • #3


Q1 = C1V and Q2 = C2V
total charge: Q = Q1 + Q2 = C1V + C2V = ( C1 + C2 )V
Ceq = Q/V = C1 + C2

I know C1 and C2, but if I use Q1/C1 = Q2/C2, I have two unknowns, both Q's. Don't I? Or can I solve for them using one of the above equations? I know Qnet, does that help??
 
  • #4


The potential difference across the 5μF capacitor is clear -- it's fixed by the 9V battery.

For the series capacitors, consider that they must both have the same amount of charge on them (any current that pushes a charge onto one of them must push the same charge onto the other because they are connected in series) and this will also be the same amount of charge that's stored on the equivalent capacitance of the pair.
 
  • #5


Find the net capacitance and thus the net charge ...

then proceed by Q1 = C1V and Q2 = C2V

and a hint, for questions like these don't find answer in terms of variables but find the value of each variable at each step ... that'll make your problem less confusing ...
 

1. What is the difference between electric potential and electric field?

Electric potential is a scalar quantity that measures the amount of electric potential energy that a charged particle has at a certain point in an electric field. It is measured in volts (V). On the other hand, electric field is a vector quantity that describes the force exerted on a charged particle by an electric field. It is measured in newtons per coulomb (N/C).

2. How do capacitors store electric charge?

Capacitors consist of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied across the plates, one plate accumulates positive charge and the other accumulates negative charge. This separation of charge creates an electric field between the plates, allowing the capacitor to store electric charge.

3. What is the relationship between capacitance and electric potential?

Capacitance is a measure of a capacitor's ability to store electric charge and is directly proportional to the electric potential difference between the plates and inversely proportional to the amount of charge stored. In other words, as the potential difference increases, the capacitance also increases, and vice versa.

4. Can capacitors store an infinite amount of charge?

No, capacitors have a maximum capacitance determined by the distance between the plates, the area of the plates, and the dielectric constant of the insulating material. Once the maximum capacitance is reached, any additional charge will cause the voltage across the plates to increase.

5. How do capacitors affect the flow of current in a circuit?

Capacitors act as temporary storage for electric charge, so they can affect the flow of current in a circuit. When a capacitor is fully charged, it will block the flow of current, acting as an open circuit. However, when a capacitor is discharged, it allows current to flow through, acting as a closed circuit. This can be useful in smoothing out fluctuations in a circuit's current or voltage.

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