Parallel Plates and Electric Field

In summary, the electrostatic potential energy of a system of point charges with charges of 1 µC, 2 µC, and 3 µC at the corners of an equilateral triangle with side length 30 cm is 0.3296 J. The electric field at a point midway between two parallel plates with a potential difference of 360 V and a distance of 0.40 cm apart is 90,000 N/C.
  • #1
Soaring Crane
469
0

Homework Statement



The electrostatic potential energy of a system of point charges q1 = 1 µC, q2 = 2 µC, and q3 = 3 µC at the corners of an equilateral triangle whose side s = 30 cm is


a. 1.10 J

b. 0.990 J

c. 0.631 J

d. 0.330 J

e. 0.123 J

Homework Equations



See below.

The Attempt at a Solution



PE = k*q*q_0/(r)

PE_total = [(k*(1*10^-6 C)*(3*10^-6 C)/(.30 m)] + [(k*(1*10^-6 C)*(2*10^-6 C)/(.30 m)] + [(k*(3*10^-6 C)*(2*10^-6 C)/(.30 m)]

= k/.30 m*[(3*10^-12 C^2) + (2*10^-12) + (6*10^-12)] = 0.3296 J ??

Homework Statement

Two flat parallel plates are a distance d = 0.40 cm apart. The potential difference between the plates is 360 V. The electric field at a point midway between the plates is approximately


a. 90 kN/C

b. 3.6 kN/C

c. 0.9 kN/C

d. zero

e. 3.6*10^5 N/C

Homework Equations



See below.

The Attempt at a Solution



E = delta V/d = 360 V/0.004 m = 90,000 V/m = 90,000 N/C ?

Thanks.
 
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  • #2
Your solutions appear correct to me.

Potential energy is the sum of the energys required to add each charge in sequence (when dealing with discrete charges). You have followed this rule in your calculations adding first the 1 microC, then the 3 microC then the 2 microC (this is implicit in the way you wrote out the equations). The order is not important since E fields superpose linearly (ie they add up in the simplest way).
The second problem also seems right to me. This is a uniform field problem and the calculated value of field strength is constant thoughout the region.
 
  • #3


Your solution for the first problem is correct. The answer is d. 0.330 J.

For the second problem, your solution is also correct. The answer is a. 90 kN/C. Keep in mind that kN stands for kilonewtons, not kilonewton per coulomb (kN/C). The correct unit for electric field is N/C. So the answer can also be written as 90,000 N/C. Great job!
 

What is the concept of electric field between parallel plates?

The concept of electric field between parallel plates refers to the force experienced by an electrically charged particle placed between two parallel plates with opposite charges. This force is created by the interaction between the electric charges on the plates and the electric charge of the particle.

How is the electric field between parallel plates calculated?

The electric field between parallel plates can be calculated by dividing the magnitude of the charge on one plate by the distance between the plates. This value is then multiplied by a constant, called the permittivity of free space, which is a measure of how easily electric fields can pass through a vacuum.

What is the direction of the electric field between parallel plates?

The electric field between parallel plates always points from the positively charged plate to the negatively charged plate. This means that the electric field lines are parallel to the plates and perpendicular to the surface of the plates.

How does the distance between parallel plates affect the electric field?

The electric field between parallel plates is inversely proportional to the distance between the plates. This means that as the distance between the plates increases, the electric field decreases and vice versa. This relationship is known as the inverse square law.

What are some real-life applications of parallel plates and electric field?

Parallel plates and electric field are used in a variety of real-life applications, such as in capacitors, which are used to store energy in electronic devices. They are also used in particle accelerators, where charged particles are accelerated through a series of parallel plates. Additionally, parallel plates and electric field are used in electrostatic air filters, which remove dust and other particles from the air using an electric field.

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