Potential Difference in a Capacitor

[cowmoo32]

Homework Statement

An isolated large-plate capacitor (not connected to anything) originally has a potential difference of 1030 volts with an air gap of 2 mm. Then a plastic slab 1 mm thick, with dielectric constant 5.1, is inserted into the middle of the air gap as shown in Figure 16.70. As shown in the diagram, location 1 is at the left plate of the capacitor, location 2 is at the left edge of the plastic slab, location 3 is at the right edge of the slab, and location 4 is at the right plate of the capacitor. All of these locations are near the center of the capacitor. Calculate the following potential differences.
V1 - V2 =
V2 - V3 =
V3 - V4 =
V1 - V4 =

Homework Equations

deltaV = E (deltaL)

The Attempt at a Solution

I assumed that the equation for deltaV would work for this problem, but apparently not. I am completely lost right now. I thought that V1-V2 would be (1030/5.1)(.0005) Where am I going wrong?

 

[cowmoo32]

anyone?

 

[ok123jump]

cowmoo32,

Let's review what we have:

$$\Delta V = 1030 = \Delta V_{1 \rightarrow 4}$$

$$K = 5.1$$

$$s_{tot} = .002m$$

$$s_{1} = .0005m$$

$$s_{2} = .001m$$

$$s_{3} = .0005m$$

with,

$$\Delta V = E \cdot s$$

and

$$\Delta V_{tot} = \sum^n_{i=1} \Delta V_i$$

One accounts for the dielectric constant of the insulator in the following fashion:

$$\Delta V_{insulator} = \frac{\Delta V_{space}}{K}$$.

To get the $$\Delta V_{1 \rightarrow 2}$$, we recognize that

$$s_{1} = .0005m = \frac{s_{tot}}{4}$$

Hence,

$$\Delta V_{1 \rightarrow 2}= E \cdot s_{1} = E \cdot \frac{s_{tot}}{4} = \frac{\Delta V}{4}$$

That should be enough information to give you the correct answer. Make sure to account for the dielectric constant of the insulator.

- Happy Problem Solving -

 

[ok123jump]

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