Parallel Plates, Capacitor and Dielectrics

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Homework Help Overview

The discussion revolves around a physics problem involving a parallel plate capacitor, where the original poster describes the setup with specific parameters such as area, separation distance, and voltage. The scenario includes the insertion of a dielectric, which affects the potential difference while keeping the charge constant.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the original capacitance and charge using provided equations but expresses uncertainty about their understanding and the correctness of their answers. Some participants question the validity of the calculations and offer encouragement to explore further.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on how to approach the next steps, particularly regarding the capacitance after the dielectric is inserted. There is a recognition of the need for clarification on certain concepts, but no consensus has been reached on the overall solution.

Contextual Notes

The original poster mentions limitations in their resources, specifically that their handouts do not cover the relevant chapter on electric fields, which may impact their understanding of the problem.

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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