# Parallel Plates, Capacitor and Dielectrics

1. Dec 3, 2015

### XYTRIX

1. The problem statement, all variables and given/known data
The parallel Plates of a capacitor have an area of 2.00 x 10-1m2 a separation distance of 1.00 X 10-2 m and are connected to a 3000 V power supply.
The capacitor is then connected from the supply, and a dielectric is inserted between the plates. We find that the potential difference decreases to 1000 V while the charge on each plate remains constant.

Find the following:
a. original capacitance
b. magnitude of charge
c. capacitance, C after a dielectric is inserted
d. the dielectric constant k of the dielectric
e. permittivity of the dielectric

2. Relevant equations
a. C=ε0 A/D
b. Q = CΔV ---im not sure

sadly the handouts given to me ends on that chapter on computing magnitude of electric field.

3. The attempt at a solution
a. C=(8.85X10-12 C2(Nm2))(2.00X10-2 m2)/(1.00X10-2 m) = 1.77X10-11
F=1.77 μF

b. Q=CΔV = (1.77X10-11 F)(3000V) = 5.31X10-8 C

this is all i know please help, also please explain things tome or did i messed this up? i know there is google but i really have a lot of things to do. please help.

2. Dec 3, 2015

### SammyS

Staff Emeritus
Hello XYTRIX. Welcome to PF !

You should easily find the answer to (c) . You know the charge, (It hasn't changed.) and you know the new potential difference.

3. Dec 3, 2015

### XYTRIX

i'm very bad at this so my answers at a and b are both right?

4. Dec 3, 2015

### SammyS

Staff Emeritus
They look right to me.