# Electrical potential help

1. Sep 12, 2005

### jpjp

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?

2. Sep 12, 2005

### FluxCapacitator

1) Here are some hints:

First, the voltage of a capacitor is equal the the charge divided by the capacitance, or $$V=Q/C$$
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: $$C=k \epsilon _0 A/d$$ 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. Sep 12, 2005

### jpjp

i get the second problem and thanks for explanations. but for some reason i still cant figure out the first one.

4. Sep 12, 2005

### Kazza_765

For the first question, put the second equation fluxC gave you into the first one ie.
$$V=\frac{Qd}{k\epsilon_0A}$$
Now everything there stays the same except V and k so you could rewrite the equation as
$$V=\frac{C}{k}$$
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.