Question about electrical potential energy

In summary, two charged plates with a total charge of 3 coulombs and a potential difference of 6 volts have a spacing of 20 cm between them. A positive charge of 0.4 coulombs is placed at the midpoint between the plates. To find its potential energy, we can use the equation E = Vq, where V is the potential at the midpoint and q is the charge. The potential at the midpoint is half of the potential difference between the plates, so the potential energy is 3 volts multiplied by 0.4 coulombs, which equals 1.2 joules. Alternatively, we can use the electric field strength between the plates, which is constant, to calculate
  • #1
Faiq
348
16

Homework Statement


Two charged plate hold a charge of 3 coulombs with the upper plate being positively charged and the lower plate being negatively charged. They have a pd of 6 volts. There is a spacing of 20 cm between them. A positive charge q with a charge 0.4 coulomb is located at midpoint. What is its potential energy?

Homework Equations


E = kQ1q/r for upper plate
E = kQ2q/r for lower plate
E = vq

The Attempt at a Solution


I tried using the first equation
E = kq/r (Q1+Q2)
E = kq/r (3-3)
E = 0 Which makes perfect sense since the lower plate is negatively charged and upper plate is positively charged

If I use E =vq
E=vq
E= 6*0.4
E = 2.4

However the answer is 1.2
What am I doing wrong
 
Physics news on Phys.org
  • #2
The charge is located at the midpoint between the plates. If we set our zero reference at the negative plate, what's the potential (electric potential) at the midpoint?
 
  • #3
Distance =2
E = 1/2 kq (Q1 +Q2)
E = 1/2kq (0)
E = 0
 
  • #4
gneill said:
The charge is located at the midpoint between the plates. If we set our zero reference at the negative plate, what's the potential (electric potential) at the midpoint?
Can you tell me the flaw in the abovementioned reasoning?
 
  • #5
Faiq said:
Can you tell me the flaw in the abovementioned reasoning?
Do you mean the part where you attempt to sum the potentials due to the charges on the plates using Coulomb's law for potential?
 
  • #6
yes
 
  • #7
Maybe I think I am taking the distance wrong. The displacement should have alternating signs right? But then how should I proceed
 
  • #8
Okay, a couple of things are problematical about that approach. First, Coulomb's law applies to point charges (or charge distributions that can be treated as point charges). Such charges spread their field lines out radially, so the inverse square law holds. The plates will have their charge spread out over a rectangular surface, and between the plates the field lines will be parallel (to a good approximation) everywhere but near the edges (fringe effect). The field due to a sheet of charge is not the same as that due to a point charge.

Second, the potential at some distance from a sheet of charge is different too, since the field lines don't diverge. The electric field is in fact uniform between the plates (excluding fringe effects again). So you can calculate the potential as the work that needs to be done to move the charge through half the plate separation (from where it is to one of the plates). Force multiplied by distance would be w = qEd, where d is 20/2 = 10 cm.

Faiq said:
Maybe I think I am taking the distance wrong. The displacement should have alternating signs right? But then how should I proceed
Electric potential doesn't depend upon direction (it's a scalar value not a vector) so the distance is not signed in the calculation.

As for the electric field strength between the plates, you're given the plate separation and the electric potential difference. The electric field has units V/m or N/C. V/m is practical here: 6 V/0.20 m = 30 V/m
 
  • #9
Oh thank you very much that was very helpful.
A little while I developed another approach, can you verify it?
E = V/d
Since E remains constant, doubling the distance (from the negative plate), will halve the potential. Thus V = 1.2
 
  • #10
Faiq said:
Oh thank you very much that was very helpful.
A little while I developed another approach, can you verify it?
E = V/d
Since E remains constant, doubling the distance (from the negative plate), will halve the potential. Thus V = 1.2
I'm not sure what distance is being doubled here, presumably the distance from the negative plate? Will your method work if the charge is placed at 1/3 of the distance of the plate separation rather than half?

The most straightforward approach is to find the electric potential (Volts) at the given location, then multiply that potential by the charge value placed there (q). The midpoint between the plates is halfway from 0 V to 6 V, so the potential is 3 V. Multiply 3 V by the charge 0.4 C.
 
  • #11
Can't I use the knowledge that V1/d1 = constant = V2/d2 in parallel plates to work out a feasible solution for such problems?
 
  • #12
Faiq said:
Can't I use the knowledge that V1/d1 = constant = V2/d2 in parallel plates to work out a feasible solution for such problems?
Yes. V/d is the electric field strength between the plates. So the electric potential at the halfway point is (V/d)(d/2) = V/2. Multiply by the charge q to get the electric potential energy.

You could also go back to the definition of potential energy and calculate the work required to move the charge from the zero potential reference (the negative plate) to the mid point. The electric force on the charge is f = qE = qV/d. The work is f ⋅ s where s is the distance the charge is moved, and equals d/2 in this case.
 
  • #13
Okay thank you very much. One more thing if the distance was 1/3 of the separation (the question you posted in your previous post) the answer would be 2V right?
 
  • #14
Faiq said:
Okay thank you very much. One more thing if the distance was 1/3 of the separation (the question you posted in your previous post) the answer would be 2V right?
The electric potential would be 2 V at that location. Multiply by the charge to get the potential energy of that charge at that location.
 
  • #15
Okay got it thank you
 

FAQ: Question about electrical potential energy

What is electrical potential energy?

Electrical potential energy is the energy stored in an object or system as a result of the separation of positive and negative charges. It is a form of potential energy that can be converted into other forms, such as kinetic energy, when the charges are allowed to move.

How is electrical potential energy different from electrical potential?

Electrical potential energy is a property of a system, while electrical potential is a measure of the potential energy per unit charge at a specific point in space. In other words, electrical potential energy is the total amount of energy stored in a system, while electrical potential is the amount of energy per unit charge at a specific location.

What factors affect the amount of electrical potential energy in a system?

The amount of electrical potential energy in a system is affected by the magnitude of the charges, the distance between them, and the medium in which they are located. The greater the magnitude of the charges and the closer they are together, the higher the electrical potential energy will be. The type of medium also plays a role, as some materials can store more electrical potential energy than others.

How is electrical potential energy calculated?

The formula for calculating electrical potential energy is U = k(q1q2)/r, where U is the potential energy, k is a constant known as the Coulomb's constant, q1 and q2 are the magnitudes of the two charges, and r is the distance between them. This equation can also be rearranged to solve for the magnitude of one of the charges if the other values are known.

What are some real-life applications of electrical potential energy?

Electrical potential energy plays a crucial role in many aspects of our daily lives, such as in the functioning of batteries, generators, and electric motors. It is also important in the transmission and distribution of electricity through power lines. Additionally, many electronic devices, such as computers and smartphones, rely on the conversion of electrical potential energy into other forms of energy to operate.

Back
Top