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Chapter 17: Problem 40

Point \(P\) is at a potential of \(500.0 \mathrm{kV}\) and point \(S\) is at a potential of \(200.0 \mathrm{kV} .\) The space between these points is evacuated. When a charge of \(+2 e\) moves from \(P\) to \(S\), by how much does its kinetic energy change?

Short Answer

Expert verified
Answer: The change in kinetic energy of the charged particle is \(9.6\times10^{-14}\,\text{J}\).

Step by step solution

01

Understand the given parameters

We are given the following parameters: - Point P's potential: \(V_P = 500.0\,\text{kV}\) - Point S's potential: \(V_S = 200.0\,\text{kV}\) - Charge moving from P to S: \(q = +2e\) The electric potential difference between the two points can be found by subtracting the potential at Point S from the potential at Point P, which gives us \(\Delta V = V_P - V_S\).
02

Calculate the electric potential difference

We can now calculate the electric potential difference: \(\Delta V = V_P - V_S = (500.0 - 200.0)\,\text{kV} = 300.0\,\text{kV} = 3 \times 10^5\,\text{V}\)
03

Calculate the potential energy change

The change in the potential energy is given by the formula: \(\Delta U = q\Delta V\) Now, substitute the values of the charge (\(q = +2e = +2 \times 1.6\times10^{-19}\,\text{C}\)) and the electric potential difference (\(\Delta V = 3 \times 10^5\,\text{V}\)) to find the potential energy change: \(\Delta U = (+2\times1.6\times10^{-19}\,\text{C})(3\times10^5\,\text{V}) = 9.6\times10^{-14}\,\text{J}\) Since the charge is positive and moving from higher potential to a lower potential, it loses potential energy.
04

Apply the work-energy principle

The work-energy principle states that the change in kinetic energy is equal to the work done on a particle. In this case, the work done on the particle is equal to the change in potential energy: \(\Delta K = \Delta U\)
05

Find the change in kinetic energy

Using the result of the potential energy change: \(\Delta K = 9.6\times10^{-14}\,\text{J}\) The kinetic energy of the particle increases by \(9.6\times10^{-14}\,\text{J}\) when it moves from point P to point S.

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Most popular questions from this chapter

(a) If the bottom of a thundercloud has a potential of $-1.00 \times 10^{9} \mathrm{V}\( with respect to Earth and a charge of \)-20.0 \mathrm{C}$ is discharged from the cloud to Earth during a lightning strike, how much electric potential energy is released? (Assume that the system acts like a capacitoras charge flows, the potential difference decreases to zero.) (b) If a tree is struck by the lightning bolt and \(10.0 \%\) of the energy released vaporizes sap in the tree, about how much sap is vaporized? (Assume the sap to have the same latent heat as water.) (c) If \(10.0 \%\) of the energy released from the lightning strike could be stored and used by a homeowner who uses \(400.0 \mathrm{kW}\). hr of electricity per month, for how long could the lightning bolt supply electricity to the home?
Figure 17.31 b shows a thundercloud before a lightning strike has occurred. The bottom of the thundercloud and the Earth's surface might be modeled as a charged parallel plate capacitor. The base of the cloud, which is roughly parallel to the Earth's surface, serves as the negative plate and the region of Earth's surface under the cloud serves as the positive plate. The separation between the cloud base and the Earth's surface is small compared to the length of the cloud. (a) Find the capacitance for a thundercloud of base dimensions \(4.5 \mathrm{km}\) by \(2.5 \mathrm{km}\) located \(550 \mathrm{m}\) above the Earth's surface. (b) Find the energy stored in this capacitor if the charge magnitude is \(18 \mathrm{C}\).
Two parallel plates are \(4.0 \mathrm{cm}\) apart. The bottom plate is charged positively and the top plate is charged negatively, producing a uniform electric field of \(5.0 \times 10^{4} \mathrm{N} / \mathrm{C}\) in the region between the plates. What is the time required for an electron, which starts at rest at the upper plate, to reach the lower plate? (Assume a vacuum exists between the plates.)
A point charge \(q=-2.5 \mathrm{nC}\) is initially at rest adjacent to the negative plate of a capacitor. The charge per unit area on the plates is $4.0 \mu \mathrm{C} / \mathrm{m}^{2}\( and the space between the plates is \)6.0 \mathrm{mm} .$ (a) What is the potential difference between the plates? (b) What is the kinetic energy of the point charge just before it hits the positive plate, assuming no other forces act on it?
The capacitor of Problem 79 is initially charged to a \(150-\mathrm{V}\) potential difference. The plates are then physically separated by another \(0.750 \mathrm{mm}\) in such a way that none of the charge can leak off the plates. Find (a) the new capacitance and (b) the new energy stored in the capacitor. Explain the result using conservation of energy.
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