Electric Potential Inside a Parallel-Plate Capacitor

In summary, the conversation is about determining the speed of two positive beads that are released from rest and very far apart. The beads have different masses and charges, with a third bead being placed exactly halfway between them. The equation used to solve for the speed is based on the conservation of energy, with the initial kinetic energy being equal to the final potential energy. The resulting calculation yields a speed of 9.6cm/s for the positive beads.
  • #1
k3N70n
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Thank you for taking the time to look. I think I get the basic idea here but I'm must be missing something important. Any help is greatly appreciated.

Homework Statement



Two 1.4g beads, each charged to 5.1nC , are 2.2cm apart. A 2.8g bead charged to -1.0nC is exactly halfway between them. The beads are released from rest.

What is the speed of the positive beads, in cm/s, when they are very far apart?
Express your answer using two significant figures.

Homework Equations



[tex]U_{i}+K_{i}=K_{f}+U_{f}[/tex]
[tex]U_{q_{1}+q_{2}}=\frac{1}{4\pi\epsilon_{0}}[/tex]

The Attempt at a Solution



[tex]m_{1}=0.0014 kg[/tex]
[tex]m_{2}=0.0014 kg[/tex]
[tex]m_{3}=0.0028 kg[/tex]
[tex]q_{1}=5.1*10^{-9} C[/tex]
[tex]q_{2}=5.1*10^{-9} C[/tex]
[tex]q_{3}=-1*10^{-9} C[/tex]
[tex]r_{12}=0.022 m[/tex]
[tex]r_{13}=0.011 m[/tex]

[tex]K_{i}=0[/tex] because initial velocity of all three is 0
[tex]U_{f}=0[/tex] because they end far apart --> r->[tex]\infty[/tex]

So [tex]U_{i}=K_{f}[/tex]
[tex]\Longrightarrow\frac{1}{4\pi\epsilon_{0}}\{\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}\}=\frac{m_{1}v_{f}^{2}}{2}[/tex]

[tex]\Longrightarrow\sqrt\frac{2}{4\pi\epsilon_{0}m_{1}}\{\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}\}}[/tex] (the sqrt sgn should extend to the end of the expression but I suck as LaTeX)
=0.096m/s=9.6cm/s
 
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  • #2
got it now.
 
  • #3


I appreciate your effort in attempting to solve this problem. However, I believe there may be some errors in your calculations and assumptions. First, the given equations do not seem to be relevant to the problem at hand, which is finding the speed of the positive beads when they are very far apart. Secondly, the initial kinetic energy should not be zero, as the beads are released from rest and therefore have some initial velocity. Additionally, the potential energy should not be zero when the beads are far apart, as there is still a non-zero electric potential between them.

To solve this problem, we can use the conservation of energy principle, which states that the total energy of a system remains constant. In this case, the initial potential energy will be converted into kinetic energy as the beads move apart. So we can set the initial potential energy equal to the final kinetic energy:

U_{i} = K_{f}

Using the equation for electric potential energy, we can write:

\frac{1}{4\pi\epsilon_{0}}\{\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}\} = \frac{m_{1}v_{f}^{2}}{2}

Solving for v_{f}, we get:

v_{f} = \sqrt{\frac{2}{m_{1}}\left(\frac{q_{1}q_{2}}{4\pi\epsilon_{0}r_{12}}+\frac{q_{1}q_{3}}{4\pi\epsilon_{0}r_{13}}\right)}

Plugging in the given values, we get:

v_{f} = \sqrt{\frac{2}{0.0014}\left(\frac{5.1*10^{-9}*5.1*10^{-9}}{4\pi*8.85*10^{-12}*0.022}+\frac{5.1*10^{-9}*(-1*10^{-9})}{4\pi*8.85*10^{-12}*0.011}\right)}

Simplifying this, we get:

v_{f} = 5.6 cm/s

Therefore, the speed of the positive beads when they are very far apart is approximately 5.6 cm/s. I hope this helps clarify any confusion and aids in your understanding
 

FAQ: Electric Potential Inside a Parallel-Plate Capacitor

1) What is a parallel-plate capacitor and how does it work?

A parallel-plate capacitor is a device used to store electric charge. It consists of two parallel plates made of conductive material with an insulating material, called a dielectric, in between. When a voltage is applied to the plates, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them. This electric field stores the electric potential energy.

2) How is the electric potential inside a parallel-plate capacitor calculated?

The electric potential inside a parallel-plate capacitor is calculated using the formula V = Ed, where V is the potential difference (voltage), E is the electric field strength, and d is the distance between the plates. This formula assumes that the electric field is constant and uniform between the plates.

3) How does the distance between the plates affect the electric potential inside a parallel-plate capacitor?

The electric potential inside a parallel-plate capacitor is directly proportional to the distance between the plates. As the distance between the plates increases, the electric potential decreases. This means that the closer the plates are together, the stronger the electric field and the higher the electric potential.

4) Can the dielectric material affect the electric potential inside a parallel-plate capacitor?

Yes, the dielectric material between the plates can affect the electric potential inside a parallel-plate capacitor. The dielectric constant of the material determines how much the electric field is reduced, and therefore, how much the electric potential is affected. A higher dielectric constant will result in a lower electric potential, while a lower dielectric constant will result in a higher electric potential.

5) How is the electric potential inside a parallel-plate capacitor related to the capacitance?

The electric potential inside a parallel-plate capacitor is directly proportional to the capacitance. The capacitance is a measure of the ability of the capacitor to store charge, and a higher electric potential results in a higher capacitance. This relationship is expressed by the formula C = Q/V, where C is the capacitance, Q is the charge stored on the plates, and V is the electric potential.

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