Find Dielectric Constant Given Charges

In summary, the presence of a dielectric slab in a parallel-plate capacitor increases the charge on each plate by 190 μC, resulting in a dielectric constant of k = 190/130 = 1.46. This can be calculated using the equation k = Q/Q0, where Q is the charge with the dielectric and Q0 is the charge without the dielectric. The capacitance equation for a parallel-plate capacitor is C = kε0A/d, where A is the area of the plates and d is the distance between them. However, since the voltage remains constant, we can use the equation Q = CV to solve for k.
  • #1
Runaway
48
0

Homework Statement


When a certain air-filled parallel-plate capacitor is connected cross a battery, it acquires a charge (on each plate) of 130 μC. While the battery connection is maintained, a dielectric slab is inserted into and fills the region between the plates. This results in the accumulation of an additional charge of 190 μC on each plate.
What is the dielectric constant of the dielectric slab?

Homework Equations



k = E0/E

The Attempt at a Solution


I have no idea, any guidance as to where to start would be great.
 
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  • #2
What's the equation for the capacitance of a parallel-plate capacitor? How does this depend on the dielectric constant?
 
  • #3
C= k[tex]\epsilon[/tex]0 A / d
so k= C/ ([tex]\epsilon[/tex]0 A / d)
But I still don't have A or d, so how does that help me?
 
  • #4
Runaway said:
C= k[tex]\epsilon[/tex]0 A / d
so k= C/ ([tex]\epsilon[/tex]0 A / d)

It should be [tex]\epsilon[/tex], not [tex]\epsilon[/tex]0, if a dielectric is present. The dielectric constant is just k=[tex]\epsilon[/tex]/[tex]\epsilon[/tex]0.

But I still don't have A or d, so how does that help me?
Well, you know that Q=CV for both situations: with the dielectric and without. You also know that V is a constant.
 
  • #5
ok, so k = q/q0
 

Related to Find Dielectric Constant Given Charges

1. How do you calculate the dielectric constant given charges?

To calculate the dielectric constant, also known as the relative permittivity, given charges, you will need to use the formula εr = C/C0, where εr is the dielectric constant, C is the capacitance with the dielectric material present, and C0 is the capacitance without the dielectric material present. You will also need to know the value of the permittivity of free space, ε0, which is 8.85 x 10^-12 F/m.

2. What is the significance of the dielectric constant in a material?

The dielectric constant is a measure of how well a material can store electrical energy in an electric field. It is an important property in electrical and electronic applications, as it affects the capacitance and impedance of a material. A high dielectric constant indicates a material can store more energy, while a low dielectric constant indicates a material that does not store as much energy.

3. How do you experimentally determine the dielectric constant of a material?

To experimentally determine the dielectric constant of a material, you will need to perform a capacitance measurement. This involves measuring the capacitance of a capacitor with and without the presence of the dielectric material. By comparing the two values, you can calculate the dielectric constant using the formula εr = C/C0. This method is known as the parallel plate method.

4. How does temperature affect the dielectric constant of a material?

The dielectric constant of a material is affected by temperature. In most materials, an increase in temperature leads to a decrease in dielectric constant. This is because as temperature increases, the thermal energy causes the molecules in the material to vibrate more, disrupting the alignment of the dipoles and reducing the material's ability to store electrical energy.

5. What are some common materials with high dielectric constants?

Some common materials with high dielectric constants include ceramic materials such as barium titanate, polymers like polyethylene and Teflon, and liquid materials such as water and sulfur hexafluoride. These materials are often used in electrical and electronic applications due to their ability to store large amounts of electrical energy.

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