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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