Question about electrical potential energy

[Faiq]

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

 

[gneill]

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?

 

[Faiq]

Distance =2
E = 1/2 kq (Q1 +Q2)
E = 1/2kq (0)
E = 0

 

[Faiq]

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?

 

[gneill]

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?

 

[Faiq]

yes

 

[Faiq]

Maybe I think I am taking the distance wrong. The displacement should have alternating signs right? But then how should I proceed

 

[gneill]

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.

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

 

[Faiq]

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

 

[gneill]

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.

 

[Faiq]

Can't I use the knowledge that V₁/d₁ = constant = V₂/d₂ in parallel plates to work out a feasible solution for such problems?

 

[gneill]

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.

 

[Faiq]

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?

 

[gneill]

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.

 

[Faiq]

Okay got it thank you