Force between capacitor plates

Click For Summary
SUMMARY

The introduction of a dielectric between capacitor plates affects the force between them. According to the discussion, while a physics textbook states that the force remains constant, the consensus among participants is that the force decreases due to a reduction in the electric field (E) caused by the dielectric material, where the permittivity (ε) is greater than 1. The surface charge density (D) is defined by the equation D = εE, indicating that for a constant charge, the electric field diminishes, leading to a decrease in force. Clarification on the circuit configuration, such as whether the capacitor is connected to a voltage source or charged and then disconnected, is crucial for accurate analysis.

PREREQUISITES
  • Understanding of capacitor fundamentals
  • Knowledge of electric fields and permittivity
  • Familiarity with the relationship between charge, electric field, and force
  • Basic concepts of circuit configurations (voltage source vs. isolated capacitor)
NEXT STEPS
  • Study the effects of dielectrics on capacitor performance
  • Learn about the equations governing electric fields in capacitors
  • Investigate the implications of circuit configurations on capacitor behavior
  • Explore advanced topics in electrostatics and material properties
USEFUL FOR

Physics students, electrical engineers, and educators seeking a deeper understanding of capacitor behavior in the presence of dielectrics.

Prateek Kumar Jain
Messages
8
Reaction score
2
What will happen to the force between the capacitor plates (Will increase or decrease or remain constant) if we introduce a dielectric between them?

As per the answer given the physics textbook , the force will remain same? But as per my understanding the force should decrease since electrical field between the plates reduces.
 
Physics news on Phys.org
The surface charge on a plate is given by the normal ##D=\epsilon E## field. The force is the charge times the ##E## field. For the same amount of charge, the ##E## field is reduced since ##\epsilon>1## so the force will go down.
 
Prateek Kumar Jain said:
As per the answer given the physics textbook , the force will remain same?
I can suggest that either the answer you have found is 'in the back of the book' and is wrong due to a simple typo or, if there is a worked example, there may be more to the question than you have stated.
For instance, what is the actual circuit? Is the Capacitor attached to a voltage source or was it charged and then disconnected?
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 103 ·
4
Replies
103
Views
9K