Dielectrics and Capacitors, Very confused.

[ryan8642]

Homework Statement

A 25pF parallel plate capacitor with an air gap b/w the plates is connected to a 100v battery. A Teflon slab is then inserted b/w the plates and completely fills the gap. What is the change in charge on the positive plate when the Teflon is inserted?

Homework Equations

Q=CV
C=eA/d
C=keA/d

e=permittivity constant

The Attempt at a Solution

I am very confused and don't know what to do, none of the formulas make sense for this question... :(
But

C=25x10^-12F
Kteflon=3.0
permittivity constant=8.85x10^-12
v=100v

what do i do? soo confused

 

[cupid.callin]

find the two capacitances after and before the plate is inserted

find the charge using Q=CV as you wrote

then just subtract the charge!

 

[ryan8642]

i don't get it, how do i find the capacitance when the teflon is inserted? do i just multiply the C (up there) by 2.0?

ktef should be 2.0 not 3.0

 

[berkeman]

i don't get it, how do i find the capacitance when the teflon is inserted? do i just multiply the C (up there) by 2.0?

ktef should be 2.0 not 3.0

Yes. By 2.0

 

[rwisz]

As a scientist, it is important to approach problems with a clear and logical thought process. In this case, we are dealing with dielectrics and capacitors, which are commonly used in electronic circuits. A capacitor is a device that stores electrical charge, and dielectrics are insulating materials that are placed between the plates of a capacitor to increase its capacitance.

To solve this problem, we need to use the equation Q=CV, where Q is the charge on the capacitor, C is the capacitance, and V is the voltage. We also need to consider the equation for capacitance of a parallel plate capacitor, which is C=eA/d, where e is the permittivity constant, A is the area of the plates, and d is the distance between the plates.

In this problem, we are given the capacitance (C=25pF) and the voltage (V=100V) of the capacitor. We also know that the Teflon slab has been inserted between the plates, completely filling the gap. This means that the distance between the plates has decreased from its original value, and the area of the plates remains the same.

To find the new capacitance, we can use the equation C=keA/d, where k is the dielectric constant of the material. In this case, the dielectric constant of Teflon is given as 3.0. Therefore, the new capacitance (C') can be calculated as C'=3.0(25pF)=75pF.

Now, using the equation Q=CV, we can find the new charge on the capacitor when the Teflon is inserted. The charge on the positive plate will be equal to the charge on the negative plate, so we can simply divide the total charge by 2. Therefore, the new charge on the positive plate (Q') can be calculated as Q'=75pF(100V)/2=3750pC.

In summary, when the Teflon slab is inserted between the plates, the charge on the positive plate increases to 3750pC. This is because the dielectric material increases the capacitance of the capacitor, allowing it to store more charge at the same voltage. I hope this explanation helps to clarify the problem and the use of the equations involved.