Calculate E and D for Parallell Plates w/ 0.5cm Plastic Plate

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In summary, the problem involves two parallel plates with a distance of 1 cm and a potential difference of 10kV. A 0.5 cm thick plastic plate with a relative permitivitty of 6 is placed between the plates. The electric field and electric displacement are the same for air and the plastic plate. The correct answer for D is 1.5e-5, but the attempted solution of D = 8.85e-6 is incorrect. The problem can be solved by treating it as two capacitors in series and calculating the electric flux D from there.
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Homework Statement



I have two parallell plates, they are separated a distance of 1 cm. The potential difference is 10kV. A 0.5 cm thick plastic plate lies between the two plates and has a relative permitivitty of 6. Calcualte E and D in the air and in the plastic plate.

Homework Equations





The Attempt at a Solution



D is the same for air and the plastic plate

The electric field without the plastic plate is [tex] E = \frac{p_s}{e_o} [/tex] and [tex] D = p_s [/tex] but [tex] E = \frac{V}{d} [/tex] that means that [tex] P_s = \frac{V*e_o}{d} = D [/tex] but that's not the right answer for D.

I get D = 8.85e-6 but it should be 1.5e-5
 
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  • #2
You can treat this as two capacitors in series: one for the air part and one for the plastic part.
 
  • #3
hmm ok..but how do I start whit the calculations then? and isn't it possible to calculate the electric flux D and go from there?
 

1. What is the formula for calculating E and D for parallel plates with a 0.5cm plastic plate?

The formula for calculating the electric field (E) and the electric displacement (D) between parallel plates with a 0.5cm plastic plate is:

E = V/d and D = εE, where V is the voltage applied to the plates, d is the distance between the plates, and ε is the permittivity of the plastic material.

2. How do you determine the voltage and distance between the plates in this calculation?

The voltage and distance between the plates can be determined by measuring the applied voltage using a voltmeter and the distance between the plates using a ruler or caliper.

3. What is the significance of the plastic plate in this calculation?

The plastic plate serves as a dielectric material between the parallel plates, which affects the electric field and electric displacement values. It can also increase the capacitance of the system.

4. Can this calculation be applied to other types of dielectric materials?

Yes, this calculation can be applied to other types of dielectric materials by using the appropriate permittivity value for the specific material in the formula.

5. How is this calculation useful in real-world applications?

This calculation is useful in determining the strength of electric fields and electric displacement in parallel plate capacitors, which are commonly used in electronic devices such as computers and smartphones. It can also be used in the study of dielectric materials and their properties.

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