Electric circuit - Why is the switch needed in this capacitor circuit?

AI Thread Summary
The discussion addresses the role of a switch in a capacitor circuit, emphasizing that a capacitor only charges when the switch is closed due to its ability to separate charge. Before the switch is closed, the capacitor does not charge significantly because the switch has an extremely small capacitance, preventing any meaningful charge transfer. The battery does not supply electrons directly but rather facilitates their movement, which is limited by the switch's properties. Additionally, the potential difference between two points in the circuit is generally consistent, with only a negligible voltage drop across the connecting wires. Understanding these principles clarifies the function of switches and capacitors in electric circuits.
Sogan
Messages
3
Reaction score
0
electric circuit -- Why is the switch needed in this capacitor circuit?

Hi,

I have two questions with respect to the picture in the attachment.:

1)
I have read, that the capacitor become charged, when the switch is closed. All right, but why the capacitor doesn't became charged before the switch is closed?
At least the left conductor is connected with the positive pole of the battery, so I thought the electrons will leave the left conductor.

2)
In the picture we can see a potential difference V between two points of the electric circuit. Is the potential V always the same between two points, whereas one Point is between the left conductor and the "+"Pole and the other one is between the right conductor and the "-"Pole?

Regards!
 

Attachments

  • Schaltkreis.jpg
    Schaltkreis.jpg
    4.4 KB · Views: 517
Physics news on Phys.org
Welcome to PF!

Hi Sogan! Welcome to PF! :smile:
Sogan said:
1)
I have read, that the capacitor become charged, when the switch is closed. All right, but why the capacitor doesn't became charged before the switch is closed?
At least the left conductor is connected with the positive pole of the battery, so I thought the electrons will leave the left conductor.

actually, it is, very very slightly …

a switch is just a capacitor with a very small plate area (A) and a very large separation (d)

since capacitance is 8.85 10-12 A/d, that means that the capacitance of a switch is extremely small (well under 10-15 F)!`

(actually the A/d formula really only applies if A is much larger than d2, but you get the idea)

capacitance is the ability to separate charge, and a switch can separate hardly any charge

if a switch and an ordinary capacitor are in series, the charge across both must be the same

for all practical purposes, the charge (across both the capacitor and the switch) can be considered zero :wink:

why can't the battery "pump" electrons onto the nearer plate of the capacitor?

a battery is electrically neutral, it doesn't supply electrons, it can only shunt them … it would have to take an electron from the switch, and the battery simply doesn't have enough strength to do that! :smile:
2)
In the picture we can see a potential difference V between two points of the electric circuit. Is the potential V always the same between two points, whereas one Point is between the left conductor and the "+"Pole and the other one is between the right conductor and the "-"Pole?

more or less …

the only difference would be the (extremly small) voltage drop across the extra bit of wire :wink:
 
Thread 'Inducing EMF Through a Coil: Understanding Flux'
Thank you for reading my post. I can understand why a change in magnetic flux through a conducting surface would induce an emf, but how does this work when inducing an emf through a coil? How does the flux through the empty space between the wires have an effect on the electrons in the wire itself? In the image below is a coil with a magnetic field going through the space between the wires but not necessarily through the wires themselves. Thank you.
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (Second part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8. I want to understand some issues more correctly. It's a little bit difficult to understand now. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. In the page 196, in the first paragraph, the author argues as follows ...
Back
Top