Parallel Plates, Capacitor and Dielectrics

In summary: You can check by seeing if the product of V and C is the same in the two cases. It should be.In summary, the parallel plates of a capacitor have an area of 2.00 x 10-1m2 and a separation distance of 1.00 X 10-2 m, connected to a 3000 V power supply. After inserting a dielectric, the potential difference decreases to 1000 V while the charge on each plate remains constant. The original capacitance is 1.77 μF and the magnitude of charge is 5.31X10-8 C. To find the new capacitance, we can use the equation C=ε0 A/D. The dielectric constant
  • #1
XYTRIX
2
0

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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  • #2
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.
 
  • #3
i'm very bad at this so my answers at a and b are both right?
 
  • #4
XYTRIX said:
I'm very bad at this so my answers at a and b are both right?
They look right to me.
 

1. What is a parallel plate capacitor?

A parallel plate capacitor is a type of capacitor that consists of two parallel conductive plates separated by a small distance. It is used to store electric charge and energy.

2. How does a parallel plate capacitor work?

A parallel plate capacitor works by creating an electric field between the two plates. When a voltage is applied, one plate becomes positively charged and the other becomes negatively charged. This creates a potential difference between the plates, allowing the capacitor to store energy.

3. What is the role of dielectrics in a parallel plate capacitor?

Dielectrics are insulating materials used between the two plates of a parallel plate capacitor. They serve to increase the capacitance of the capacitor by reducing the electric field between the plates. This allows for a greater charge to be stored.

4. How do you calculate the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor can be calculated by using the equation C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates.

5. What are some common uses of parallel plate capacitors?

Parallel plate capacitors have a variety of uses in electronic circuits, including power supplies, filtering circuits, and energy storage in devices like cameras and flashlights. They are also used in high voltage applications, such as in particle accelerators and plasma generators.

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