Finding electric potential difference

In summary, the conversation discusses the determination of potential difference between two parallel plates with equal and opposite surface charge densities. The problem is approached by using the equation V=Ed and assigning an arbitrary area to the plates. It is found that the voltage remains the same regardless of the area, but the charge on the plates changes. This method allows for the calculation of the voltage without knowing the dimensions of the plates.
  • #1
wendo
15
0

Homework Statement



Two parallel plates having charges of equal magnitude but opposite sign are separated by 14.2 cm. Each plate has a surface charge density of 33.3 nC/ m2. A proton is released from rest at the positive plate. Determine the potential difference between the plates.


Homework Equations


E=surface charge density/epsilon knot (8.85X10-12)

V=Ed

The Attempt at a Solution


I tried to relate the electric field equation to voltage difference by V=Ed but this doesn't seem to work. Is there something that I'm missing?

Here are my two approraches:

1st:

Electric field=n/epsilon knot= 3.33x10-9/8.85x10-12
=376.3

V=Ed
=752.6(0.142m)
=106.8C/m

This was wrong so then i tried to find q by solving it from E=kq/r^2
and then using the q to find what potential difference would be through the equation electric potential=kq/r

so from there I got electric potential is 53.43 and since we need to find the difference of voltage i did:

delta v= 53.43- (-53.43)
=106.8 the same thing as before!

What am I missing and what am I doing wrong?? I only have 2 tries left :(
 
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  • #2
Well, one way to do it would be to assign an arbitrary area to the plates, and solve for the voltage. Then change the area of the plates, and notice that the voltage... does what?

Try this: set the parallel plates to a size of 1m X 1m, so the charge on each plate is 33.3nC. Now you know that Q=CV, and you have an equation that you can use for the capacitance of a parallel plate capacitor in terms of d, A, and epsilon, correct? What voltage do you calculate?

Now try the same thing with the plates twice as big (4x area)...what voltage do you get?
 
  • #3
wendo said:
1st:

Electric field=n/epsilon knot= 3.33x10-9/8.85x10-12
=376.3

V=Ed
=752.6(0.142m)
=106.8C/m
Looks like you doubled the field from one formula to the next, for some reason.

This was wrong so then i tried to find q by solving it from E=kq/r^2
and then using the q to find what potential difference would be through the equation electric potential=kq/r
Those formulas are for point charges--nothing to do with this problem.
 
  • #4
Using an arbitrary area worked! THANK YOU SOOOOOO MUCH!
(i'm going to warn you now that i might have a lot of questions today...)

So the voltage will be the same regardless of what the area is, but in order to keep it the same the charge on the plates will change. I see!

I was just really confused at how to be able to find out what the charge was from a charge density without knowing the dimensions, so now I know just to assume a 1m^2 area!

Thank youuu!
 

1. What is electric potential difference?

Electric potential difference is the difference in electric potential between two points in an electric field. It is a measure of the energy required to move a unit charge from one point to another.

2. How is electric potential difference calculated?

Electric potential difference is calculated by dividing the change in electric potential energy by the amount of charge moved. This can be represented by the formula V = ΔU/q, where V is the electric potential difference, ΔU is the change in electric potential energy, and q is the amount of charge.

3. What is the unit of measurement for electric potential difference?

The unit of measurement for electric potential difference is the volt (V). It is equivalent to one joule of energy per coulomb of charge.

4. How is electric potential difference related to electric field?

Electric potential difference is directly related to electric field. The electric field is the force per unit charge at a given point, while electric potential difference is the energy per unit charge. Therefore, the electric field can be calculated by dividing the electric potential difference by the distance between two points.

5. What factors affect the electric potential difference?

The electric potential difference is affected by the amount of charge, the distance between two points, and the strength of the electric field. It is also affected by the medium through which the charge is moving, as different materials have different resistances to the flow of electricity.

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