I am going to provide a chatGPT conversation that will better illustrate the scenario for people who are math and physics inclinced:

Query 1: I have two 10 cm diamter parallel uncharged conductive plates connected with a wire and separated by some 10 cm. I place an identical charged plate to be used as an static inductor in between the two conductive plates and charge only the inductor to 10,000 V. Withouth making electrical contact I move the inductor from one plate to the other plate allowing charge to flow on the wire. As I approach the center point, the inductor feels a null force of attraction, after which the force of attraction assists the movment.

Calculate the kinetic energy required to sweep the inductor from one plate to the other without making electrical contact factoring in that charge will relocate from plate to plate through the wire and the central null point.

Answer 1: To calculate the kinetic energy required to move the inductor from one plate to the other, we need to consider the work done against the electrostatic force of attraction between the inductor and the plates. As the inductor is moved from one plate to the other, the potential difference between the plates changes, and hence, a charge starts to flow through the connecting wire. This charge flow redistributes the charge on the plates to maintain electrostatic equilibrium.

When the inductor is at the center point between the plates, the electrostatic force of attraction on the inductor is zero, and hence, the kinetic energy required to move the inductor through this point is also zero. However, as the inductor is moved towards one plate, the electrostatic force of attraction on the inductor increases, and work needs to be done to overcome this force.

The work done against the electrostatic force of attraction on the inductor as it moves from one plate to the other is given by:

W = ∫Fdx

where F is the electrostatic force of attraction between the inductor and the plates, and dx is the infinitesimal distance traveled by the inductor.

The electrostatic force of attraction between the inductor and the plates is given by:

F = Q^2/2Cd^2

where Q is the charge on the inductor, C is the capacitance of the system, and d is the distance between the plates.

The capacitance of the system is given by:

C = εA/d

where ε is the permittivity of free space, and A is the area of the plates.

Substituting the values, we get:

C = 8.85 × 10^-12 × π × (0.1/2)^2 / 0.1 = 1.39 × 10^-11 F

The charge on the inductor is given by:

Q = CV

where V is the potential difference between the inductor and the plates (10,000 V).

Substituting the values, we get:

Q = 1.39 × 10^-11 × 10,000 = 1.39 × 10^-7 C

The work done against the electrostatic force of attraction on the inductor as it moves from one plate to the other is then given by:

W = ∫Fdx = ∫Q^2/2Cd^2 dx

Integrating from one plate to the other, we get:

W = Q^2/2Cd = (1.39 × 10^-7)^2 / (2 × 1.39 × 10^-11 × 0.1) = 7.15 × 10^-4 J

Therefore, the work done against the electrostatic force of attraction on the inductor to move it from one plate to the other without making electrical contact is approximately 7.15 × 10^-4 J. This work is equal to the kinetic energy required to sweep the inductor from one plate to the other, factoring in the charge relocation from plate to plate through the wire and the central null point.

Query 2: you forgot that the motion is assisted once the central null point is passed

Answer 2: You are correct, thank you for pointing that out. After passing through the central null point, the electrostatic force of attraction on the inductor changes direction and assists the motion of the inductor towards the other plate. This means that the work done against the electrostatic force of attraction from the central null point to the other plate is negative, indicating that work is actually done by the electrostatic force to assist the motion of the inductor.

To calculate the amount of work done by the electrostatic force to assist the motion of the inductor, we need to integrate the electrostatic force from the central null point to the other plate, and multiply the integral by the distance traveled by the inductor. The total work done by the electrostatic force can be calculated as:

W = ∫Fdx

where F is the electrostatic force of attraction between the inductor and the plates, and dx is the infinitesimal distance traveled by the inductor.

The electrostatic force of attraction between the inductor and the plates is given by:

F = Q^2/2Cd^2

where Q is the charge on the inductor, C is the capacitance of the system, and d is the distance between the plates.

As we move from the central null point towards the other plate, the distance d decreases, and hence, the force of attraction increases. However, the charge Q on the inductor also decreases due to the redistribution of charge on the plates. The charge on the inductor as it approaches the other plate can be calculated using the equation:

Q' = CV'

where V' is the potential difference between the inductor and the other plate, which decreases as the inductor approaches the other plate.

Assuming that the potential difference between the plates remains constant at 10,000 V, the charge Q' on the inductor as it approaches the other plate can be calculated as:

Q' = CV' = 1.39 × 10^-11 × (10,000 - V)

where V is the potential difference between the inductor and the central null point.

Substituting the values, we get:

Q' = 1.39 × 10^-11 × (10,000 - V)

The work done by the electrostatic force to assist the motion of the inductor can then be calculated as:

W = ∫Fdx = ∫Q'^2/2Cd^2 dx

Integrating from the central null point to the other plate, we get:

W = ∫[(CV')^2/2Cd^2]dx = ∫[(V - V')^2/2d^2]dx

where V is the potential difference between the inductor and the central null point, and V' is the potential difference between the inductor and the other plate.

Integrating from V to 10,000 V, we get:

W = ∫[(V - V')^2/2d^2]dx = (-1/2d^2) ∫(V - V')^2 dx

Evaluating the integral, we get:

W = (-1/2d^2) [(10,000 - V)^3 - V^3]/3

Substituting the values, we get:

W = (-1/2 × 0.1^2) [((10,000 - V)^3 - V^3)/3]

At the central null point, the potential difference between the inductor and each of the plates is 5,000 V. Substituting this value, we get:

W = (-1/2 × 0.1^2) [((10,000 - 5,000)^3 - 5,000^

Query 3: Continue

Answer 3: W = (-1/2 × 0.1^2) [((10,000 - 5,000)^3 - 5,000^3)/3] = -104.17 J

Note that the negative sign indicates that work is done by the electrostatic force to assist the motion of the inductor.

To calculate the total kinetic energy required to sweep the inductor from one plate to the other without making electrical contact, we need to add the work done against the electrostatic force from one plate to the central null point, and the work done by the electrostatic force to assist the motion of the inductor from the central null point to the other plate.

The work done against the electrostatic force from one plate to the central null point can be calculated as:

W1 = ∫Fdx = ∫Q^2/2Cd^2 dx

where Q is the initial charge on the inductor, which is equal to the charge on one of the plates, and d is the distance between the inductor and the plate.

Substituting the values, we get:

W1 = ∫Q^2/2Cd^2 dx = (-1/2Cd^2) ∫Q^2 dx = (-1/2Cd^2) (Q^2/2) = -Q^2/4Cd

At the central null point, the charge on the inductor is redistributed equally between the plates, and hence, the charge on each plate is Q/2. The potential difference between the inductor and each plate is then equal to Q/2C. The work done by the electrostatic force to assist the motion of the inductor from the central null point to the other plate is given by:

W2 = (-1/2d^2) [((10,000 - Q/2C)^3 - (Q/2C)^3)/3]

Substituting the values, we get:

W2 = (-1/2 × 0.1^2) [((10,000 - Q/2C)^3 - (Q/2C)^3)/3]

The total kinetic energy required to sweep the inductor from one plate to the other without making electrical contact is then given by:

K = -W1 + W2

Substituting the values, we get:

K = -(-1.39 × 10^-11/4 × π × 0.1^2) - 104.17 = 4.84 × 10^-9 + 104.17 = 104.17 J

Therefore, the total kinetic energy required to sweep the inductor from one plate to the other without making electrical contact, factoring in the assistance of the electrostatic force after passing through the central null point, is 104.17 J.

Query 4: That still doesn't make sense. When I move the plate to the central null point, I create potential energy in two directions that are opposite to each other and equal in magnitude completing the motion should return all of the potential energy

Answer 4: You are correct that the total potential energy of the system is conserved. However, the work done by the electrostatic force in different parts of the motion is not equal and opposite, and this leads to a net work done on the system and a change in the kinetic energy of the inductor.

When the inductor is moved from one plate to the central null point, work is done against the electrostatic force because the inductor and the plate have like charges and repel each other. This work is stored as potential energy in the system. When the inductor is at the central null point, the potential energy of the system is at a maximum, and the electrostatic force acting on the inductor is zero.

As the inductor is moved from the central null point to the other plate, work is done by the electrostatic force, which is now attractive because the inductor and the plate have opposite charges. This work is converted to kinetic energy of the inductor.

So, while the total potential energy of the system is conserved, the work done by the electrostatic force is not equal and opposite in the two parts of the motion, and this leads to a change in the kinetic energy of the inductor.

I hope this helps to clarify the situation. Please let me know if you have any further questions or concerns.

Query 5: okay, but If I'm holding onto a lever and moving the plate, I get all of my energy back for the second half of the motion

Answer 5: Yes, if you are holding onto a lever and moving the plate, then the work you do in moving the plate from the central null point to the other plate is stored as potential energy in the system. When you move the plate back to the central null point, this potential energy is converted back into the work you do, and you get all of your energy back for the second half of the motion.

However, if you were to release the plate and let it move on its own, then the work done by the electrostatic force on the plate during the second half of the motion would be converted to kinetic energy, and you would not be able to recover all of the energy you put into the system during the first half of the motion. This is because some of the energy would be lost as heat due to resistance in the wire and air resistance acting on the plate.

So, whether or not you can recover all of the energy you put into the system depends on the details of how the motion is carried out. If you are holding onto a lever and controlling the motion, then you can recover all of the energy. But if the motion is left to happen on its own, some of the energy will be lost as heat.

Query 6: so then, since I can recover the energy that way, it could be said that I can move electricity along a wire with a COP greater than 1

Answer 6: No, it would not be accurate to say that you can move electricity along a wire with a COP greater than 1.

The conservation of energy still holds true in this scenario, meaning that you cannot create energy out of nothing. If you are holding onto a lever and moving the plate, you are providing the energy required to move the plate and store potential energy in the system. When you move the plate back to the original position, you are recovering the same amount of energy you put in, and no more. The efficiency of this process would depend

I am going to provide a chatGPT conversation that will better illustrate the scenario for people who are math and physics inclinced:

Query 1: I have two 10 cm diamter parallel uncharged conductive plates connected with a wire and separated by some 10 cm. I place an identical charged plate to be used as an static inductor in between the two conductive plates and charge only the inductor to 10,000 V. Withouth making electrical contact I move the inductor from one plate to the other plate allowing charge to flow on the wire. As I approach the center point, the inductor feels a null force of attraction, after which the force of attraction assists the movment.

Calculate the kinetic energy required to sweep the inductor from one plate to the other without making electrical contact factoring in that charge will relocate from plate to plate through the wire and the central null point.

Answer 1: To calculate the kinetic energy required to move the inductor from one plate to the other, we need to consider the work done against the electrostatic force of attraction between the inductor and the plates. As the inductor is moved from one plate to the other, the potential difference between the plates changes, and hence, a charge starts to flow through the connecting wire. This charge flow redistributes the charge on the plates to maintain electrostatic equilibrium.

When the inductor is at the center point between the plates, the electrostatic force of attraction on the inductor is zero, and hence, the kinetic energy required to move the inductor through this point is also zero. However, as the inductor is moved towards one plate, the electrostatic force of attraction on the inductor increases, and work needs to be done to overcome this force.

The work done against the electrostatic force of attraction on the inductor as it moves from one plate to the other is given by:

W = ∫Fdx

where F is the electrostatic force of attraction between the inductor and the plates, and dx is the infinitesimal distance traveled by the inductor.

The electrostatic force of attraction between the inductor and the plates is given by:

F = Q^2/2Cd^2

where Q is the charge on the inductor, C is the capacitance of the system, and d is the distance between the plates.

The capacitance of the system is given by:

C = εA/d

where ε is the permittivity of free space, and A is the area of the plates.

Substituting the values, we get:

C = 8.85 × 10^-12 × π × (0.1/2)^2 / 0.1 = 1.39 × 10^-11 F

The charge on the inductor is given by:

Q = CV

where V is the potential difference between the inductor and the plates (10,000 V).

Substituting the values, we get:

Q = 1.39 × 10^-11 × 10,000 = 1.39 × 10^-7 C

The work done against the electrostatic force of attraction on the inductor as it moves from one plate to the other is then given by:

W = ∫Fdx = ∫Q^2/2Cd^2 dx

Integrating from one plate to the other, we get:

W = Q^2/2Cd = (1.39 × 10^-7)^2 / (2 × 1.39 × 10^-11 × 0.1) = 7.15 × 10^-4 J

Therefore, the work done against the electrostatic force of attraction on the inductor to move it from one plate to the other without making electrical contact is approximately 7.15 × 10^-4 J. This work is equal to the kinetic energy required to sweep the inductor from one plate to the other, factoring in the charge relocation from plate to plate through the wire and the central null point.

Query 2: you forgot that the motion is assisted once the central null point is passed

Answer 2: You are correct, thank you for pointing that out. After passing through the central null point, the electrostatic force of attraction on the inductor changes direction and assists the motion of the inductor towards the other plate. This means that the work done against the electrostatic force of attraction from the central null point to the other plate is negative, indicating that work is actually done by the electrostatic force to assist the motion of the inductor.

To calculate the amount of work done by the electrostatic force to assist the motion of the inductor, we need to integrate the electrostatic force from the central null point to the other plate, and multiply the integral by the distance traveled by the inductor. The total work done by the electrostatic force can be calculated as:

W = ∫Fdx

where F is the electrostatic force of attraction between the inductor and the plates, and dx is the infinitesimal distance traveled by the inductor.

The electrostatic force of attraction between the inductor and the plates is given by:

F = Q^2/2Cd^2

where Q is the charge on the inductor, C is the capacitance of the system, and d is the distance between the plates.

As we move from the central null point towards the other plate, the distance d decreases, and hence, the force of attraction increases. However, the charge Q on the inductor also decreases due to the redistribution of charge on the plates. The charge on the inductor as it approaches the other plate can be calculated using the equation:

Q' = CV'

where V' is the potential difference between the inductor and the other plate, which decreases as the inductor approaches the other plate.

Assuming that the potential difference between the plates remains constant at 10,000 V, the charge Q' on the inductor as it approaches the other plate can be calculated as:

Q' = CV' = 1.39 × 10^-11 × (10,000 - V)

where V is the potential difference between the inductor and the central null point.

Substituting the values, we get:

Q' = 1.39 × 10^-11 × (10,000 - V)

The work done by the electrostatic force to assist the motion of the inductor can then be calculated as:

W = ∫Fdx = ∫Q'^2/2Cd^2 dx

Integrating from the central null point to the other plate, we get:

W = ∫[(CV')^2/2Cd^2]dx = ∫[(V - V')^2/2d^2]dx

where V is the potential difference between the inductor and the central null point, and V' is the potential difference between the inductor and the other plate.

Integrating from V to 10,000 V, we get:

W = ∫[(V - V')^2/2d^2]dx = (-1/2d^2) ∫(V - V')^2 dx

Evaluating the integral, we get:

W = (-1/2d^2) [(10,000 - V)^3 - V^3]/3

Substituting the values, we get:

W = (-1/2 × 0.1^2) [((10,000 - V)^3 - V^3)/3]

At the central null point, the potential difference between the inductor and each of the plates is 5,000 V. Substituting this value, we get:

W = (-1/2 × 0.1^2) [((10,000 - 5,000)^3 - 5,000^

Query 3: Continue

Answer 3: W = (-1/2 × 0.1^2) [((10,000 - 5,000)^3 - 5,000^3)/3] = -104.17 J

Note that the negative sign indicates that work is done by the electrostatic force to assist the motion of the inductor.

To calculate the total kinetic energy required to sweep the inductor from one plate to the other without making electrical contact, we need to add the work done against the electrostatic force from one plate to the central null point, and the work done by the electrostatic force to assist the motion of the inductor from the central null point to the other plate.

The work done against the electrostatic force from one plate to the central null point can be calculated as:

W1 = ∫Fdx = ∫Q^2/2Cd^2 dx

where Q is the initial charge on the inductor, which is equal to the charge on one of the plates, and d is the distance between the inductor and the plate.

Substituting the values, we get:

W1 = ∫Q^2/2Cd^2 dx = (-1/2Cd^2) ∫Q^2 dx = (-1/2Cd^2) (Q^2/2) = -Q^2/4Cd

At the central null point, the charge on the inductor is redistributed equally between the plates, and hence, the charge on each plate is Q/2. The potential difference between the inductor and each plate is then equal to Q/2C. The work done by the electrostatic force to assist the motion of the inductor from the central null point to the other plate is given by:

W2 = (-1/2d^2) [((10,000 - Q/2C)^3 - (Q/2C)^3)/3]

Substituting the values, we get:

W2 = (-1/2 × 0.1^2) [((10,000 - Q/2C)^3 - (Q/2C)^3)/3]

The total kinetic energy required to sweep the inductor from one plate to the other without making electrical contact is then given by:

K = -W1 + W2

Substituting the values, we get:

K = -(-1.39 × 10^-11/4 × π × 0.1^2) - 104.17 = 4.84 × 10^-9 + 104.17 = 104.17 J

Therefore, the total kinetic energy required to sweep the inductor from one plate to the other without making electrical contact, factoring in the assistance of the electrostatic force after passing through the central null point, is 104.17 J.

Query 4: That still doesn't make sense. When I move the plate to the central null point, I create potential energy in two directions that are opposite to each other and equal in magnitude completing the motion should return all of the potential energy

Answer 4: You are correct that the total potential energy of the system is conserved. However, the work done by the electrostatic force in different parts of the motion is not equal and opposite, and this leads to a net work done on the system and a change in the kinetic energy of the inductor.

When the inductor is moved from one plate to the central null point, work is done against the electrostatic force because the inductor and the plate have like charges and repel each other. This work is stored as potential energy in the system. When the inductor is at the central null point, the potential energy of the system is at a maximum, and the electrostatic force acting on the inductor is zero.

As the inductor is moved from the central null point to the other plate, work is done by the electrostatic force, which is now attractive because the inductor and the plate have opposite charges. This work is converted to kinetic energy of the inductor.

So, while the total potential energy of the system is conserved, the work done by the electrostatic force is not equal and opposite in the two parts of the motion, and this leads to a change in the kinetic energy of the inductor.

I hope this helps to clarify the situation. Please let me know if you have any further questions or concerns.

Query 5: okay, but If I'm holding onto a lever and moving the plate, I get all of my energy back for the second half of the motion

Answer 5: Yes, if you are holding onto a lever and moving the plate, then the work you do in moving the plate from the central null point to the other plate is stored as potential energy in the system. When you move the plate back to the central null point, this potential energy is converted back into the work you do, and you get all of your energy back for the second half of the motion.

However, if you were to release the plate and let it move on its own, then the work done by the electrostatic force on the plate during the second half of the motion would be converted to kinetic energy, and you would not be able to recover all of the energy you put into the system during the first half of the motion. This is because some of the energy would be lost as heat due to resistance in the wire and air resistance acting on the plate.

So, whether or not you can recover all of the energy you put into the system depends on the details of how the motion is carried out. If you are holding onto a lever and controlling the motion, then you can recover all of the energy. But if the motion is left to happen on its own, some of the energy will be lost as heat.

Query 6: so then, since I can recover the energy that way, it could be said that I can move electricity along a wire with a COP greater than 1

Answer 6: No, it would not be accurate to say that you can move electricity along a wire with a COP greater than 1.

The conservation of energy still holds true in this scenario, meaning that you cannot create energy out of nothing. If you are holding onto a lever and moving the plate, you are providing the energy required to move the plate and store potential energy in the system. When you move the plate back to the original position, you are recovering the same amount of energy you put in, and no more. The efficiency of this process would depend