• Support PF! Buy your school textbooks, materials and every day products Here!

Capacitor change in distance

  • Thread starter jaydnul
  • Start date
  • #1
516
11

Homework Statement


A parallel plate capacitor with plates of area [itex]460 cm^2[/itex] is charged to a potential difference [itex]V[/itex] then disconnected from the source of voltage. When the plates move [itex].3 cm[/itex] farther apart, the voltage between them increases by [itex]120 V[/itex]. What is the charge [itex]Q[/itex] on the positive plate of the capacitor? Answer in [itex]nC[/itex]


Homework Equations


[tex]Q=\frac{ε_0AV}{d}[/tex]


The Attempt at a Solution


[tex]V=\frac{Qd}{ε_0A}[/tex]
[tex]Q=\frac{ε_0A(V+120)}{(d+.003)}[/tex]
Inserting one into the other...
[tex]Q=\frac{ε_0A}{(d+.003)}(\frac{Qd}{ε_0A}+120)[/tex]
distribute and canceling yields...
[tex]Q=\frac{Qd+120ε_0A}{(d+.003)}[/tex]
Finally
[tex]Q=\frac{120ε_0A}{.003}[/tex]

So using [itex]A=.046 m^2[/itex], I keep getting [itex]16nC[/itex]

Am I doing something wrong assuming [itex]Q[/itex] is constant?
 

Answers and Replies

  • #2
gneill
Mentor
20,792
2,770
Your assumption about Q being constant is correct, and I don't see any errors in your method. That leaves significant figures or application errors (the marking program has been given a wrong answer to look for) as suspects.
 
  • #3
516
11
Ok thanks for the feedback. This homework system has a reputation for making errors.
 

Related Threads on Capacitor change in distance

Replies
4
Views
3K
Replies
1
Views
4K
Replies
17
Views
501
  • Last Post
Replies
4
Views
1K
Replies
9
Views
5K
Replies
2
Views
1K
  • Last Post
Replies
4
Views
7K
Replies
5
Views
1K
Replies
10
Views
1K
Top