Exploring Electrical Voltage U between Parallel Flat Plates

[mmoadi]

Homework Statement

- How much is the electrical voltage U between two evenly electrified parallel flat plates, distanced d = 1 cm, if the strength of electric field between them is E = 1 V / m?
- What is the strength of electric field to the left and to the right of each plate?
- What is the force on the surface (area) unit between the two plates?
- What are the voltage between the plates and the force on the surface unit as a result of doubling the distance between the two plates?
- How is the strength of the electric field and force between the two plates altered (changed), if the voltage is maintained constant?
Influencing constant is ε(0) = 8.85 × 10^-12 As / Vm.

Homework Equations

V= E*d
F= Q*E
V= Qd/ Aε(0)
E= Q/ Aε(0)
Q= CV

The Attempt at a Solution

First part: How much is the electrical voltage U between two evenly electrified parallel flat plates?

V= 1 V/m* 0.01 m= 0.01 V

Second part: What is the strength of electric field to the left and to the right of each plate?

E= V/ d
E= 0.01 v/ 0.01 m= 1 V/m

Third part: What is the force on the surface unit between the two plates?

E= ½ Q*V → Q= 2E/ V
Q= 200 C

F= Q*E= 200 N

Fourth part: What are the voltage between the plates and the force on the surface unit as a result of doubling the distance between the two plates?

d= 2 cm= 0.02 m

V= d*E= 0.02 V
Q= 2E/ V= 100 C
F= Q*E= 100 N

ARE MY CALCULATIONS CORRECT?

Fifth part: How is the strength of the electric field and force between the two plates altered (changed), if the voltage is maintained constant?

I need some help with this part. Any hints?

Thank you for helping!

 

[willem2]

The second part: The field between the plates is 1 V/m, but the field outside the plates isn't

Third part: The E in $E = \frac{1}{2}QV$ is energy, the E in $F = Q E$ is the strength of the electric field, so this calculation is wrong. You need the force per square meter anyway. It's easiest to assume the area of the capacitor is 1 $m^2$ and then to compute the capacity and the charge and then use $F = Q E$

Fourth part: wrong for the same reason as third part.

Fifth part: if the voltage is the same, the electric field strength, which is volt/meter must change
if the distance changes. use V = E * d.

 

[mmoadi]

_{Thank you for helping, I will just have to do additional calculating!}