How Is Capacitance Calculated in a Charged Parallel-Plate Capacitor?

Click For Summary
SUMMARY

The capacitance of a charged parallel-plate capacitor is calculated using the formula C = Q/V, where Q is the charge and V is the potential difference. In this discussion, the charges on the plates are +900uC and +100uC, resulting in a potential difference of 4V. The correct calculation for capacitance involves the total charge transferred, which is 400uC, leading to a capacitance of 100uC. Understanding the relationship between charge transfer and capacitance is crucial for accurate calculations.

PREREQUISITES
  • Understanding of capacitance and the formula C = Q/V
  • Familiarity with electric charge units, specifically microcoulombs (uC)
  • Basic knowledge of potential difference in electrical circuits
  • Introduction to Gauss' law and its relevance in electrostatics
NEXT STEPS
  • Study the derivation and applications of the capacitance formula C = Q/V
  • Learn about the principles of Gauss' law and its implications for electric fields
  • Explore the effects of charge distribution on capacitance in different geometries
  • Investigate the behavior of capacitors in AC circuits and their role in filtering
USEFUL FOR

Students studying physics, electrical engineers, and anyone interested in understanding the principles of capacitance and electrostatics.

crossfacer
Messages
20
Reaction score
0

Homework Statement


The two plates of a parallel-plate capacitor initially carry equal amount of positive charge. If some charges are transferred from one plate to another, the charges on the plates are respectively +900uC and +100uC. The potential difference across the plates becomes 4V. What is the capacitance of the capacitor?

Homework Equations



C=Q/V

The Attempt at a Solution


I just pick randomly pick +900uC instead of +100uC:
C=900/4=225uC

4. Then I check the answer:

Totally 400uC moves away from one to the metal plate and move to another plate, so:
C=400/4=100uC

According to what I have learn, Q is the charge of one of the metal plate. In here, why should I subsitute 400uC, the amount of charge that is transferred?
 
Physics news on Phys.org
Consider the relative charge...

Have you met Gauss' law?
 
Last edited:
I haven't heard about it before. I will try to learn it by myself. Thank you:rolleyes:
 

Similar threads

Replies
11
Views
3K
Replies
6
Views
867
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 14 ·
Replies
14
Views
1K
Replies
11
Views
3K
Replies
4
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
1
Views
1K
Replies
5
Views
963