Physics true/false questions v. isolated parallel plates

Click For Summary

Homework Help Overview

The discussion revolves around true/false questions related to the behavior of isolated parallel plates with opposite charges, focusing on capacitance, stored energy, and the effects of distance and dielectrics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationships between distance, capacitance, stored energy, and the effects of inserting a dielectric. Some express uncertainty about specific statements, while others attempt to clarify the implications of formulas related to capacitance and energy.

Discussion Status

Some participants have provided guidance on the relationships between capacitance and distance, as well as the role of dielectrics. Multiple interpretations of the statements are being explored, with some participants affirming or questioning the correctness of various assertions.

Contextual Notes

Participants are working under the constraint that the plates are isolated and cannot lose charge, which influences their reasoning about the effects of distance and dielectrics on capacitance and energy.

Gonger
Messages
10
Reaction score
0
Which of the statements below are true for two oppositely charged, isolated parallel plates? (C is the capacitance, U is the stored energy, +Q and -Q are the charges on the plates.) Note: isolated plates can not lose their charge. (Enter ALL correct statements, e.g., BCD)

A) When the distance is doubled, U increases.
True. I am thinking that if the distance is doubled there is more to store energy in.

B) Inserting a dielectric increases C.
I have no idea on this one.

C) When the distance is halved, Q stays the same.
True. I don't think distance will affect the charge.

D) Increasing the distance increases the electric field.
True.

E) When the distance is doubled, C increases.
False. It doesn't have any affect on the capacitance

F) Inserting a dielectric decreases U.
Again no idea.

G) Inserting a dielectric increases Q.
I just have no clue with dielectrics.

Thanks for any help in advance.
 
Physics news on Phys.org
Some forumlas you need to know is

[tex]C = \frac {Q}{V} = \epsilon_0 \frac {A}{d}, \ \ U = \frac {Q^2}{2C}[/tex]

So when d increases, C decreases, then U increase.

When ever you are dealing with a dielectric, the capacitance is given by

[tex]C_d = \kappa \epsilon_0 \frac {A}{d}[/tex]

Where [itex]\kappa[/itex] is a constant called the dielectric constant. This is always greater than 1. This says that [itex]C_d > C[/itex]. Which is exactly why dielectrics are inserted, to increase the capacitance.

That should get you started.
 
cool. I got it thanks.
 
Gonger said:
B) Inserting a dielectric increases C.
True. [itex]C \propto \epsilon > \epsilon_0[/itex]

D) Increasing the distance increases the electric field.
True.
False. The field is uniform between the plates and is independent of the separation where the plate size is large compared to separation. The potential depends upon distance (V = Ed)

E) When the distance is doubled, C increases.
False.
True. C is proportional to d.

F) Inserting a dielectric decreases U.
True. The dielectric reduces the field between the plates. So it reduces the potential (V=Ed),

G) Inserting a dielectric increases Q.
False. Charge is constant.

So: ABCEF

AM
 
Andrew Mason said:
True. C is proportional to d.

I think [itex]C \propto \frac {1}{d}[/itex].
 
Corneo said:
I think [itex]C \propto \frac {1}{d}[/itex].
Right you are. So ABCF.

AM
 

Similar threads

Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K