# Calculating the voltage and electric field strength between two parallel plates

In summary, the electrical voltage between two evenly electrified parallel flat plates with a distance of 1 cm and an electric field strength of 1 V/m is 0.01 V. To find the strength of electric field to the left and right of each plate, use the equation Q= CV. The force on the surface unit between the two plates is 200 N. When the distance between the plates is doubled to 2 cm, the voltage between the plates and the force on the surface unit are 0.02 V and 100 N, respectively. If the voltage is maintained constant, the strength of the electric field and force between the plates will remain the same.

## 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.

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

## 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

Are my calculations for the first part correct?

And I really don't know how to approach the rest of the problem. Can someone please help me? Any hints?

Last edited:

## 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.

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

## 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

Are my calculations for the first part correct?

And I really don't know how to approach the rest of the problem. Can someone please help me? Any hints?

Yes, V = 0.01V is correct for the first part. The other equation that you will need is Q = CV. That should give you what you need for the rest.

Thank you for helping!

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

I need some help with this part. Any hints?

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!

Last edited:

## 1. How do you calculate the voltage between two parallel plates?

To calculate the voltage between two parallel plates, you can use the formula V = Ed, where V is the voltage, E is the electric field strength, and d is the distance between the two plates. Alternatively, you can use the formula V = Q/C, where Q is the charge on the plates and C is the capacitance of the system.

## 2. What is the electric field strength between two parallel plates?

The electric field strength between two parallel plates is a measure of the force per unit charge that a test charge would experience if placed between the plates. It is typically represented by the symbol E and is measured in units of volts per meter (V/m).

## 3. How does the distance between the plates affect the voltage and electric field strength?

The distance between the plates is directly proportional to the voltage and inversely proportional to the electric field strength. This means that as the distance between the plates increases, the voltage also increases, while the electric field strength decreases.

## 4. What factors affect the electric field strength between two parallel plates?

The electric field strength between two parallel plates is affected by the magnitude and distribution of charge on the plates, as well as the distance between them. It is also influenced by the material properties of the plates and the surrounding medium, such as the dielectric constant.

## 5. Can the voltage and electric field strength between two parallel plates be manipulated?

Yes, the voltage and electric field strength between two parallel plates can be manipulated by changing the distance between the plates, altering the charge or capacitance of the system, or using different materials for the plates and surrounding medium. This is a fundamental principle in the design and operation of many electrical devices such as capacitors and parallel plate accelerators.

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