Electric Field in Dielectric: Charge Distribution & Calculations

In summary, two identical, oppositely-charged, conducting plates separated by a dielectric with a dielectric constant of 3.60 and thickness of 1.80 mm have a resultant electric field of 1.2*10^6 V/m. Using Gauss's law, the charge per unit area on the conducting plates is found to be 3.82*10^-5 C/m². To calculate the charge per unit area on the surfaces of the dielectric, the bound surface charge density is needed, which is equal to the normal component of the polarization vector P. The answer for part b is 2.76*10^-5 C/m².
  • #1
Qyzren
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Two identical, oppositely-charged, conducting plates of area 2.50cm² are separated by a dielectric 1.80 mm thick, with a dielectric constant of 3.60. The resultant electric field in the dielectric is 1.2*10^6 V/m.

a)Compute the charge per unit area on the conducting plate.
b)Compute the charge per unit area on the surfaces of the dielectric.

The Attempt at a Solution


for part a). I used gauss's law for dielectric.
Q/A = K*epsilon_0*E = 3.60*8.85*10^-12*1.2*10^6 = 3.82*10^-5 C/m² which is correct.

However i haven't got a single clue how to do part b. But the answer is suppose to be 2.76*10^-5 C/m². Can someone please help me figure out how this answer was obtained?
 
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  • #2
For b, they must mean bound surface charge density, which equals the normal component of the polarization vector P.
 
  • #3
thank you for help. i managed to figure it out now with your tip :)
 

1. What is an electric field in a dielectric?

An electric field in a dielectric is the force per unit charge experienced by a charge when placed in a dielectric material. It is caused by the presence of polar molecules or ions in the dielectric material, which can align themselves with an external electric field.

2. How is charge distribution affected by a dielectric material?

When a dielectric material is placed in an electric field, the polar molecules or ions in the material will experience a force and align themselves with the field. This results in a redistribution of charge within the material, as the positive and negative charges are separated from each other.

3. How do you calculate the electric field in a dielectric material?

The electric field in a dielectric material can be calculated using the equation E = 1/(4πε0εr) * Q/R2, where E is the electric field, ε0 is the permittivity of free space, εr is the relative permittivity of the material, Q is the charge, and R is the distance from the charge to the point where the electric field is being measured.

4. What is the role of dielectric constant in calculating the electric field?

The dielectric constant, also known as the relative permittivity, is a measure of how easily a material can be polarized by an external electric field. It affects the strength of the electric field in the material, as seen in the equation E = 1/(4πε0εr) * Q/R2. A higher dielectric constant means that the material is more easily polarized and will have a stronger electric field.

5. How does the presence of a dielectric material affect the capacitance of a capacitor?

The presence of a dielectric material between the plates of a capacitor will increase the capacitance of the capacitor. This is because the material reduces the electric field between the plates, allowing for more charge to be stored on the plates. The capacitance can be calculated using the equation C = εr * C0, where C is the new capacitance, εr is the relative permittivity of the dielectric, and C0 is the capacitance without the dielectric.

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