Two circular discs spaced apart from a parallel plate capacitor ?

In summary, the question involves finding the diameter of two circular disks in a parallel-plate capacitor with a given electric field strength caused by the transfer of electrons. The equations used are E = Q / (e_o * A) and A = [pie]*r^2. The solution involves calculating the charge and using the calculated value to find the diameter, but there was an error in the calculation of the charge leading to an incorrect diameter. Once the error is corrected, the correct diameter can be determined.
  • #1
Netsurfer733
17
0

Homework Statement



Two circular disks spaced 0.50 mm apart from a parallel-plate capacitor. Transferring 1.90 x 10^9 electrons from one disk to the other causes the electric field strength to be 1.90 x 10^5 N/C. What are the diameters of the disks (in cm)?


Homework Equations



E = Q / (e_o * A)
A = [pie]*r^2

The Attempt at a Solution



I though I went through the right process for this, but maybe I'm not using the right equations somehow? Here's what I did:

E = Q / (e_o * A) ==> Q = (8.85*10^-12 C^2 / N*m^2) * A * (1.9*10^5 N/C) = 3.04 * 10^-1

A = [Pie]*r^2 ==> 1.81 * 10^-4 = [pie]*r^2 ==> r = 0.00759 and thusly D = 0.1517m which is equal to 15.2 cm. This turned out to be wrong.

Where did I mess up?
 
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  • #2
Your derivation is a bit confusing; you seem to be mixing in pre-calculated values without showing how you got them. For example, I assume that the value "3.04 * 10^-1" is supposed to be the charge due to the moved electrons. The mantissa looks okay, but the power of ten is way off. Perhaps your value for A is off because of that.
 
  • #3
well, the first thing i see is that your diameter is more than twice your radius.

the first line of your solution is confusing; it looks like you are doing many steps at once and i am not sure what the last number is supposed to be since there are no units (if it is supposed to be the charge, it seems too big).

the reasoning is correct as far as i can tell, but you need to check your arithmetic and the value of the charge.
 
  • #4
eczeno said:
well, the first thing i see is that your diameter is more than twice your radius.


D'oh! That was it! Hahah - thanks much, and sorry for the confusion :) You just allowed me to pass this assignment ^.^
 
  • #5




Your calculations are correct, but you have made a mistake in converting the units. The given distance between the disks is 0.50 mm, which is equal to 0.005 m. Your answer for the diameter is in meters, but the question asks for it in centimeters. Therefore, the correct answer is 15.17 cm. It is always important to pay attention to units and make sure they are consistent throughout the calculation.
 

Related to Two circular discs spaced apart from a parallel plate capacitor ?

1. What is a parallel plate capacitor?

A parallel plate capacitor is a type of capacitor that consists of two conductive plates separated by an insulating material, called a dielectric. It is used to store electrical energy by creating an electric field between the plates.

2. How does the distance between the two circular discs affect the capacitance of the capacitor?

The capacitance of a parallel plate capacitor is directly proportional to the surface area of the plates and inversely proportional to the distance between them. This means that as the distance between the two circular discs increases, the capacitance decreases and vice versa.

3. What is the role of the dielectric material in a parallel plate capacitor?

The dielectric material in a parallel plate capacitor serves to increase the capacitance by reducing the electric field between the plates. It also acts as an insulator to prevent a direct flow of current between the plates.

4. How is the capacitance of a parallel plate capacitor calculated?

The capacitance of a parallel plate capacitor can be calculated using the formula C = ε0A/d, where C is the capacitance, ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

5. What are some real-world applications of a parallel plate capacitor?

Parallel plate capacitors have many practical applications, including in electronic circuits, power factor correction, energy storage systems, and high voltage power supplies. They are also used in devices such as radios, televisions, and computers.

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