Parallel Plates, Capacitor and Dielectrics

AI Thread Summary
The discussion revolves around calculating various properties of a parallel plate capacitor with a dielectric. The original capacitance is calculated as 1.77 μF, and the charge on the plates is determined to be 5.31 x 10^-8 C when connected to a 3000 V supply. After inserting the dielectric, the potential difference drops to 1000 V, prompting further calculations for the new capacitance and dielectric constant. Participants confirm that the initial calculations for capacitance and charge are correct. The thread emphasizes the importance of understanding these concepts for solving related physics problems.
XYTRIX
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Homework Statement


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

Homework 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.
i wanted to know about this badly.

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.
 
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XYTRIX said:

Homework Statement


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

Homework 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.
i wanted to know about this badly.

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.
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.
 
i'm very bad at this so my answers at a and b are both right?
 
XYTRIX said:
I'm very bad at this so my answers at a and b are both right?
They look right to me.
 

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