Question Relating Electric Field and Capacitance

In summary, the conversation discusses the calculation for the magnitude of charge on each plate of a parallel-plate capacitor with a given electric field and plate dimensions. The equation Q=epsilon(E)(A) is used to find the charge, but the individual asking the question is getting half of the correct answer. The expert suggests that a possible explanation could be the presence of a dielectric with a dielectric constant of 2, but the question did not mention this information.
  • #1
rabcdred
13
0

Homework Statement


A parallel-plate capacitor has a plate area of 0.2 m2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 × 10^6 V/m between the plates, the magnitude of the charge on each plate should be


Homework Equations



Q=epsilon(E)(A)

The Attempt at a Solution


I keep getting half of the correct answer. Does anyone know why I am off a factor of 2? Cheers.
 
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  • #2
Q=CV
C=epsilon*Area/separation
 
  • #3
rabcdred said:
I keep getting half of the correct answer. Does anyone know why I am off a factor of 2?

No, without seeing your solution.

ehild
 
  • #4
My guess would be that you are only considering the electric field caused by one of the plates. ?
 
  • #5
Without seeing your solution, my guess is that the correct answer is twice what your answer is.
 
  • #6
Sorry about that.
Q=CV, C=(epsilon)A/d, V=Ed/ By substituting these last 2 equations into the first, the result is Q=(epsilon)A(E)-->8.85e-12(0.2)(2e6)= 3.54e-6. However, this is half of the correct answer. What am I doing wrong?
 
  • #7
That is correct with the given data. But it is possible that the volume between the plates was filled by a dielectric, with dielectric constant of 2. Are you sure that the dielectric constant was not given?

ehild
 
  • #8
Yes. The question I gave you was a copy paste from the document.
 
  • #9
I have no more idea. It happens quite often that wrong solutions are given. ehild
 

1. What is an electric field?

An electric field is a physical field that surrounds a charged particle and exerts a force on other charged particles within its influence. It can be thought of as the space around a charged particle where the force of the particle's charge can be felt.

2. How is electric field related to capacitance?

Capacitance is a measure of an object's ability to store electric charge. It is directly proportional to the strength of the electric field and inversely proportional to the distance between the charged objects. In other words, the stronger the electric field and the closer the charged objects, the higher the capacitance.

3. What is the equation for calculating electric field?

The equation for calculating electric field is E = F/q, where E is the electric field strength in Newtons per Coulomb (N/C), F is the force in Newtons (N), and q is the charge in Coulombs (C).

4. How does distance affect electric field and capacitance?

As distance between charged objects decreases, the electric field strength increases and the capacitance also increases. This is because the closer the objects are, the stronger the force of the electric field and the more charge can be stored between them.

5. What are some real-world applications of electric field and capacitance?

Electric field and capacitance have many practical applications, such as in electronic devices like capacitors, which store electric charge. They are also important in the functioning of batteries, generators, and electric circuits. Additionally, understanding these concepts is crucial in fields such as electrical engineering, physics, and astronomy.

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