Answer to Question #84351 in Physics for Adam

Question #84351
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
1
Expert's answer
2019-01-21T15:47:27-0500

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

D=ϵE=ϵ0ϵrE,D = \epsilon E = \epsilon_0 \epsilon_r E,

here,

DD

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

ϵ\epsilon

is the permittivity of the dielectric,

ϵ0\epsilon_0

is the permittivity of the free space,

ϵr\epsilon_r

is the relative permittivity of the dielectric,

EE

is the electric field intensity.

Since, charge density equals flux density, we get:

E=Dϵ0ϵr.E = \frac{D}{\epsilon_0 \epsilon_r}.

Let's substitute the numbers:

E=5104μCcm2106C1μC(100cm)21m28.8541012Fm3=188238Vm.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=188238Vm0.15103m=28.2V.V = Ed = 188238 \frac{V}{m} \cdot 0.15 \cdot 10^{-3} m = 28.2 V.

Answer:

E=188238Vm,V=28.2VE = 188238 \frac{V}{m}, V = 28.2 V

.

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