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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

The inside of a cell membrane is at a potential of \(90.0 \mathrm{mV}\) lower than the outside. How much work does the electric field do when a sodium ion (Na') with a charge of \(+e\) moves through the membrane from outside to inside?
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 \(18 \mathrm{C}\).
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