Electrical potential help

  • Thread starter jpjp
  • Start date
  • #1
jpjp
2
0
electrical potential help..urgent!

1) A parallel-plate capacitor is made of two flat metal plates pressed against a thin slab of dielectric material. The capacitor is connected to a power supply, and a potential difference of 100 V is applied to the plates. With the power supply disconnected, the dielectric material is removed and the potential difference between the plates is measured to be 500 V. What is the dielectric constant of the material that was initially used to fill the gap between the plates?

2) Points A and B have electric potentials of 275 V and 129 V, respectively. When an electron released from rest at point A arrives at point C its kinetic energy is KA. When the electron is released from rest at point B, however, its kinetic energy when it reaches point C is KB = 4KA. What is the electric potential at point C?
 

Answers and Replies

  • #2
FluxCapacitator
52
0
1) Here are some hints:

First, the voltage of a capacitor is equal the the charge divided by the capacitance, or [tex]V=Q/C[/tex]
Second, charge is always conserved, in everything physics(so far at least), so that's a constant you'd want to use.
Last, the dielectric changes the capacitance according to the equation: [tex]C=k \epsilon _0 A/d[/tex] where k is the dielectric constant you're looking for. (A and d stay the same in with or without the dielectric in this case)

2) qV=energy, where q is charge and V is the potential difference. Using this equation you can find the difference in energy from the path from A to C and from B to C, and then work from there.

If you need more help than these admittedly vague hints, I'll give a solution, but it's always best to solve things for yourself.
 
  • #3
jpjp
2
0
i get the second problem and thanks for explanations. but for some reason i still can't figure out the first one.
 
  • #4
Kazza_765
171
0
For the first question, put the second equation fluxC gave you into the first one ie.
[tex]V=\frac{Qd}{k\epsilon_0A}[/tex]
Now everything there stays the same except V and k so you could rewrite the equation as
[tex]V=\frac{C}{k}[/tex]
Where C is a constant. You also know that if k=1 (ie. no dielectric present) V=500, so you can solve for C. Then let V=100 and find k.
 

Suggested for: Electrical potential help

Replies
17
Views
700
Replies
5
Views
533
Replies
1
Views
195
Replies
6
Views
335
  • Last Post
Replies
10
Views
339
Replies
11
Views
432
Replies
5
Views
333
  • Last Post
Replies
9
Views
368
Replies
1
Views
317
Top