Uniform elecric fields- force and kinetic energy

In summary, increasing the separation between conducting plates has a direct effect on both force and kinetic energy. According to F=VQ/d, as the distance increases, the force decreases. Similarly, the gain in kinetic energy, which is equivalent to work done, is also affected as it is dependent on the force. Additionally, the stored energy in a capacitor is given by W=(1/2)CV^2, where C is capacitance and V is voltage, both of which are affected by the separation distance. However, if Q remains constant, the force is not affected by the separation. This can also be seen in the equation Fx=(1/2)Q^2/e0A, where x is the separation distance
  • #1
chanderjeet
25
0
what is the effect on the force as well as the gain in kinetic energy if the separation of the conducting plates is increased?

I'm thinking that since F= VQ/d if the distance increases then the force would decrease. I could be wrong.

Since, the gain in KE is equivalent to the work done =Fd and since F decreases (I assume) then the KE would also decrease.
 
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  • #2
The stored energy in a capacitor is

W = (1/2)CV2= (1/2)Q2/C = (1/2)xQ2/e0A
where x = separation and A = area.
So Fx= (dW/dx)= (1/2)Q2/e0A
So if Q is an invariant, the force is independent of separation.
Also Fx = dW/dx = d/dx(1/2)CV2=(1/2)e0AV2 d/dx(1/x)= -(1/2)e0AV2/x2
where V is held constant.
Bob S
 
  • #3
sorry, i didn't specify. I was referring to, say, a particle being released at the upper plate of horizontal parallel conducting plates. (in a vacuum)
 
  • #4
In this case, Q=CV
so V = Q/C = Qd/e0A
where A = area and d the separation
so if Q = constant, and d is increased, the voltage between the plates increases linearly.

One way to generate high voltage between the plates is to charge them up when d is small, then separate the two plates.
Bob S
 

Related to Uniform elecric fields- force and kinetic energy

1. What is a uniform electric field?

A uniform electric field is a region in which the electric field strength and direction are constant. This means that any charged particle placed in the field will experience the same force and move in a straight line.

2. How is the force exerted by a uniform electric field calculated?

The force exerted by a uniform electric field on a charged particle is calculated by multiplying the electric field strength by the charge of the particle. This can be represented by the equation F = qE, where F is the force, q is the charge, and E is the electric field strength.

3. What is the relationship between the force and kinetic energy of a charged particle in a uniform electric field?

The force and kinetic energy of a charged particle in a uniform electric field are directly proportional. This means that as the force increases, the kinetic energy of the particle also increases. This relationship can be represented by the equation KE = qEd, where KE is the kinetic energy, q is the charge, E is the electric field strength, and d is the distance the particle travels in the field.

4. How does the direction of the force on a charged particle in a uniform electric field relate to the direction of the electric field?

The force on a charged particle in a uniform electric field is always in the same direction as the electric field. This means that if the electric field is pointing to the right, the force on a positively charged particle will also be pointing to the right. If the electric field is pointing to the left, the force on a negatively charged particle will also be pointing to the left.

5. Can a charged particle have kinetic energy in a uniform electric field even if it is not moving?

Yes, a charged particle can have kinetic energy in a uniform electric field even if it is not moving. This is because the kinetic energy is the energy of motion, and even if the particle is not moving, it still has the potential to move due to the force exerted by the electric field. This is similar to a ball sitting at the top of a hill, which has the potential to roll down and gain kinetic energy even though it is not currently moving.

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