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Chapter 17: Problem 16

An electron is moved from point \(A\), where the electric potential is \(V_{A}=-240 \mathrm{V},\) to point \(B,\) where the electric potential is \(V_{B}=-360 \mathrm{V}\). What is the change in the electric potential energy?

Short Answer

Expert verified
Answer: The change in the electric potential energy of the electron is 1.922 x 10^-17 J.

Step by step solution

01

Identify given variables

We are given: - The electric potential at point A: \(V_A = -240\:V\) - The electric potential at point B: \(V_B = -360\:V\) - The charge of an electron: \(e = -1.602 \times 10^{-19}\:C\)
02

Calculate the difference in electric potentials

To find the change in electric potential energy, we need to first calculate the difference in electric potential: \(\Delta V = V_B - V_A = (-360\: V) - (-240\: V) = -120\: V\)
03

Calculate the change in electric potential energy

Now, we will calculate the change in electric potential energy using the formula: \(\Delta U = q\Delta V\) where \(q\) is the charge of the electron and \(\Delta V\) is the difference in electric potentials. Substitute the values into the formula: \(\Delta U = (-1.602 \times 10^{-19}\:C)(-120\:V)\) Now, calculate the value: \(\Delta U = 1.922 \times 10^{-17}\:J\) The change in the electric potential energy of the electron is \(\Rightarrow \Delta U = 1.922 \times 10^{-17}\:J\).

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Most popular questions from this chapter

A cell membrane has a surface area of \(1.0 \times 10^{-7} \mathrm{m}^{2},\) a dielectric constant of \(5.2,\) and a thickness of \(7.5 \mathrm{nm}\) The membrane acts like the dielectric in a parallel plate capacitor; a layer of positive ions on the outer surface and a layer of negative ions on the inner surface act as the capacitor plates. The potential difference between the "plates" is \(90.0 \mathrm{mV}\). (a) How much energy is stored in this capacitor? (b) How many positive ions are there on the outside of the membrane? Assume that all the ions are singly charged (charge +e).
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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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