# Homework Help: Finding The Dielectric Constant

1. Mar 5, 2013

### Bashyboy

1. The problem statement, all variables and given/known data
The voltage across an air-filled parallel-plate capacitor is measured to be 181.0 V. When a dielectric is inserted and completely fills the space between the plates as in the figure below, the voltage drops to 27.0 V.

What is the dielectric constant of the inserted material?

2. Relevant equations

3. The attempt at a solution

Using the expression that relates the capacitance of capacitor before a dielectric is placed between the plates, to the capacitance of a capacitor after the dielectric is places between the plates, $C=\kappa C_0$, I solved for $\kappa$: $\kappa = \frac{C}{C_0}$. I then substituted in the expression of for capacitance: $\kappa = \frac{ \frac{Q}{\Delta V}}{ \frac{Q_0}{\Delta V_0}}$

Once I came to this step, I wasn't sure how to proceed. I wasn't sure if the charge would remain constant from each situation. It would seem like the charge wouldn't remain the same in each one, because the the amount of charge a capacitor can hold depends on the voltage across it and its capacitance, both of which are reducing.

2. Mar 5, 2013

### tia89

That depends on the set up. If you charged the plates to reach the initial voltage $\Delta V_0$ and then you isolate the plates form the rest of the world, the charge HAS to be constant, because it has nowhere to go and no way to get there (and from the data you have it seems this is what happens).

Notice that here the drop in voltage is due to the insertion of the dielectric... when you insert a dielectric the molecules inside it polarizes so as to partially screen the original field (how much depends indeed on the dielectric constant) and thus the voltage decreases.

3. Mar 5, 2013

### Bashyboy

I have another question, and it being answered might help me with this problem: If there is a potential difference between the parallel plate capacitor, is it necessary that there is a flow of charge (current)?

4. Mar 5, 2013

### tia89

You need an electric field, not necessarily a current... the current eventually establishes DUE to the electric field (or potential difference if you prefer) when there is the possibility for it to establish... meaning some conductor material.

In your case as the two plates should not be connected by some wire (also because in this case they would discharge and you wouldn't have a capacitor anymore) there is no current.