How Much Charge Is on an Isolated Spherical Conductor?

  • Thread starter Thread starter XxXDanTheManXxX
  • Start date Start date
  • Tags Tags
    Charge Test
Click For Summary

Homework Help Overview

The discussion revolves around an isolated spherical conductor charged to a specific voltage, with participants exploring the relationships between charge, voltage, and capacitance. The problem involves calculating the charge on the conductor, determining its capacitance, and examining how capacitance changes with varying voltage levels.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between charge and voltage, with some suggesting the use of equations relating capacitance and electric field. Questions arise regarding how changes in voltage affect capacitance and whether charge and voltage can be treated independently.

Discussion Status

There is an ongoing exploration of the relationships between the variables involved, with some participants providing equations and others questioning the assumptions made. While some guidance has been offered, there is no clear consensus on the interpretations of the relationships between charge, voltage, and capacitance.

Contextual Notes

Participants note that the original poster has not provided their attempts at solving the problems, leading to concerns about the nature of the homework help being sought. There is a mention of imposed homework rules regarding the need for personal effort in understanding the material.

XxXDanTheManXxX
Messages
8
Reaction score
0
Studying help

An isolated spherical conductor of radius 20cm is charged to 4k V (i.e. 4000 V).
a. How much charge is on the conductor?
b. What is the capacitance of the sphere?
c. how does the capacitance change if the spere is charged to 8 k V (8000)?
 
Physics news on Phys.org
Im not sure about a.

b) Use the relationship Q = CV to find C.
c) Use the solved equation above to find the effect on C if V = 2V
 
Since you have test tomorrow I do not know if this will be of any help, but here it is anyway.
a=R=20cm

a.)
[tex] 4000 V=- \int_\infty^a \vec{E} dr =-\int_\infty^a \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} dr[/tex]

b.)
[tex] Q= C \Delta V[/tex]
[tex] \frac{Q}{\Delta V}= C[/tex]
[tex] \Delta V =-\int_\infty^a \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} dr[/tex]
steps not shown
[tex] C=4 \pi \epsilon_0 R[/tex]

3.) What Whozum did not mention here is that Q and V are related so it is not as easy as he says it is. I provided about an equation that should work just fine for you.
 
Davorak:

With [tex]V = \frac{Q}{4\pi\epsilon_0 R} \mbox{ and } C = \frac{Q}{V} = 4\pi\epsilon_0R[/tex]

Does it not follow that

[tex]2V = \frac{Q}{2\pi\epsilon_0 R} \mbox{ so then } C = \frac{Q}{2V} = 2\pi\epsilon_0R[/tex] ?

The charges cancel in either case, no?
 
whozum said:
Davorak:

With [tex]V = \frac{Q}{4\pi\epsilon_0 R} \mbox{ and } C = \frac{Q}{V} = 4\pi\epsilon_0R[/tex]

Does it not follow that

[tex]2V = \frac{Q}{2\pi\epsilon_0 R} \mbox{ so then } C = \frac{Q}{2V} = 2\pi\epsilon_0R[/tex] ?

The charges cancel in either case, no?
You are considering Charge and Voltage independent of each other when they are not.
Take your second example:

[tex] 2V = \frac{2 Q}{4\pi\epsilon_0 R} \mbox{ so then } C = \frac{2 Q}{2V} = 4\pi\epsilon_0 R[/tex]

When you increase Q you increase V and vs versa for this problem.
 
So doubling the voltage has no effect on capacitance because doubling the voltage doubles the charge?
 
By George I think you’ve got it. Really vs versa though since charge is usually considered fundamental while voltage is not.
 
XxXDanTheManXxX said:
Studying help

An isolated spherical conductor of radius 20cm is charged to 4k V (i.e. 4000 V).
a. How much charge is on the conductor?
b. What is the capacitance of the sphere?
c. how does the capacitance change if the spere is charged to 8 k V (8000)?

You appear to have posted a SERIES of homework questions on here - many of them actually are related, meaning that if you understand how to solve one, the others should follow along the same line. However, in all of the questions you have posted, not ONCE have you even described your attempt or what you have tried in solving the problem, nor are we told what you do know. I mean, you MUST have some clue on where to start, or else we would just be doing the home work for you and doing you a complete disservice.

Please READ the STICKY for this section of PF before you post more questions.

Zz.
 
Not to disagree with ZapperZ, nevertheless this is all the OP should need

[tex]V=k_eq/r[/tex]
 
  • #10
GeneralChemTutor said:
Not to disagree with ZapperZ, nevertheless this is all the OP should need

[tex]V=k_eq/r[/tex]

And not to disagree with you either, but if the OP does not even know this, then "Houston, we have a problem!"

Zz.
 
  • #11
ZapperZ said:
GeneralChemTutor said:
Not to disagree with ZapperZ, nevertheless this is all the OP should need
[tex]V=k_eq/r[/tex]


And not to disagree with you either, but if the OP does not even know this, then "Houston, we have a problem!"

Zz.
Knowing that would certainly help him for spherical cases, but he does not seem to know how to do it general. On another thread he was unaware how to get a electric field in a parallel plate capacitor. This is why I gave the more general form. He seems to be memorizing equations rather knowing how to derive them.
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 16 ·
Replies
16
Views
7K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 8 ·
Replies
8
Views
7K
Replies
21
Views
2K
Replies
5
Views
3K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K