Parallel-Plate Capacitor Question

[Lovergoo]

Homework Statement

A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.50 cm wide and 15.0 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick.

What is the maximum charge that can be stored in this capacitor? (The dielectric constant of mica is 5.4, and its dielectric strength is 1.00* 10^8 V/m.

Homework Equations

C=(Eo*A)/d
V=Ed=(Eo/K)*d

The Attempt at a Solution

First I calculated A from the given dimensions, 0.525 m^2. Then I calculated C with [(8.85*10^-12)(0.525)]/(2.25*10^-5 m). Next, I tried to calculate V by taking the dielectric breakdown value of 1.00*10^8 and dividing by the dielectric constant 5.4. I then multiplied that value by the distance, 2.25*10^-5 m giving me 416.67 V. I multiplied V*C to find Qmax. The value is 8.6*10^-5 -- for some reason this isn't right.

Where did I go wrong?

Thanks!

 

[Vuldoraq]

Homework Equations

C=(Eo*A)/d
V=Ed=(Eo/K)*d

The Attempt at a Solution

One of these equations are not correct for a capacitor with a dielectric. They should be,

$$C=\frac{k\epsilon_{0}A}{A}$$

By $$E_{0}$$ do you mean electric field or permitivity of free space?

 

[nlightner]

Your approach seems correct, but there may be a calculation error. Here is the correct solution:

1. First, calculate the capacitance using the formula C = (ε₀A)/d, where ε₀ is the permittivity of free space (8.85*10^-12 F/m), A is the area (0.525 m^2) and d is the distance between the plates (2.25*10^-5 m). This gives a capacitance of 1.86*10^-9 F.

2. Next, calculate the maximum electric field that the mica strip can withstand using the formula E = V/d, where V is the voltage and d is the thickness of the mica strip (0.0225 mm = 2.25*10^-5 m). This gives an electric field of 1.85*10^12 V/m.

3. Since the dielectric strength of mica is 1.00*10^8 V/m, the maximum voltage that can be applied is 1.85*10^12 V/m divided by 1.00*10^8 V/m, which gives a maximum voltage of 18.5 V.

4. Finally, the maximum charge that can be stored in the capacitor is Qmax = CV = (1.86*10^-9 F)(18.5 V) = 3.44*10^-8 C.

Therefore, the maximum charge that can be stored in this capacitor is 3.44*10^-8 C.