Answer to Question #84351 in Physics for Adam

Question

A parallel plate capacitor has plate of 0.15mm apart and dielectric with relative Permittivity of 3. Find the electric field intensity and the voltage between the plate if the surface is 5*E-4 micrometer coulomb/Cm²

Answer 1

We can find the electric field intensity between the plates from the formula:

$$D = \epsilon E = \epsilon_0 \epsilon_r E,$$

here,

$$D$$

is the electric displacement field (in free space, the electric displacement field is equivalent to flux density ),

$$\epsilon$$

is the permittivity of the dielectric,

$$\epsilon_0$$

is the permittivity of the free space,

$$\epsilon_r$$

is the relative permittivity of the dielectric,

$$E$$

is the electric field intensity.

Since, charge density equals flux density, we get:

$$E = \frac{D}{\epsilon_0 \epsilon_r}.$$

Let's substitute the numbers:

$$E = \frac{5 \cdot 10^{-4} \frac{\mu C}{cm^2} \cdot \frac{10^{-6} C}{1 \mu C} \cdot \frac{(100 cm)^2}{1 m^2}}{8.854 \cdot 10^{-12} \frac{F}{m} \cdot 3} = 188238 \frac{V}{m}.$$

We can find the voltage between the plates from the formula:

$$V = Ed = 188238 \frac{V}{m} \cdot 0.15 \cdot 10^{-3} m = 28.2 V.$$

Answer:

$$E = 188238 \frac{V}{m}, V = 28.2 V$$

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