Shell in a cylindrical capacitor

Click For Summary
SUMMARY

The discussion focuses on calculating the capacitance of a cylindrical capacitor before and after the insertion of a cylindrical conducting shell. The initial capacitor consists of a conductor with radius R1 and an outer cylindrical surface with radius R2, charged to a potential V0. Participants debate the impact of electrostatic induction and the configuration of capacitors in series, concluding that the final capacitance Cfinal must be twice the initial capacitance Cinitial, necessitating a specific thickness d for the shell.

PREREQUISITES
  • Cylindrical capacitor theory
  • Capacitance calculations for series and parallel configurations
  • Understanding of electrostatic induction
  • Mathematical manipulation of capacitance formulas
NEXT STEPS
  • Calculate the capacitance of a cylindrical capacitor using the formula C = (2 * π * ε0 * L) / ln(R2/R1)
  • Explore the effects of inserting a conducting shell on capacitance
  • Learn about electrostatic induction and its implications in capacitor design
  • Investigate the conditions under which capacitance can be doubled in a cylindrical capacitor setup
USEFUL FOR

Students studying electromagnetism, electrical engineers designing capacitors, and anyone interested in advanced capacitor configurations and their electrical properties.

fcoulomb
Messages
8
Reaction score
0

Homework Statement


There is a capacitor (llenght ##L##) made of a conductor (cylinder of radius ##R_1##) and a cylindrical surface (radius ##R_2##). It is charged with a potential ##V_0##, then it is isolated.
(There is vacuum between ##R_1## and ##R_2##)

Now we insert a cylindrical conducting shell of thickness ##d##, and inside radius ##R_3##, as in the picture below.1) Find the capacitance before and after the insertion of the shell. Find the value of ##d## such that ##C_{final}=2 C_{initial}## .

The Attempt at a Solution


[/B]
My question is: after the shell is inserted, there is electrostatic induction between the shell and the ##R_1## cylinder, so i can see the problem as two capacitors in series, BUT the total final capacitance would be always less than the initial.

So how I can start?
 

Attachments

  • fotoforum.PNG
    fotoforum.PNG
    8.7 KB · Views: 375
Physics news on Phys.org
fcoulomb said:

Homework Statement


There is a capacitor (llenght ##L##) made of a conductor (cylinder of radius ##R_1##) and a cylindrical surface (radius ##R_2##). It is charged with a potential ##V_0##, then it is isolated.
(There is vacuum between ##R_1## and ##R_2##)

Now we insert a cylindrical conducting shell of thickness ##d##, and inside radius ##R_3##, as in the picture below.1) Find the capacitance before and after the insertion of the shell. Find the value of ##d## such that ##C_{final}=2 C_{initial}## .

The Attempt at a Solution


[/B]
My question is: after the shell is inserted, there is electrostatic induction between the shell and the ##R_1## cylinder, so i can see the problem as two capacitors in series, BUT the total final capacitance would be always less than the initial.

So how I can start?
You start by showing some work. Write down the capacitance before the insertion and the capacitances of the two series capacitors after the insertion. Add them as capacitors in series and see if you get a value that is less than the initial. Maybe you won't.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
7
Views
7K