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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

If a capacitor has a capacitance of \(10.2 \mu \mathrm{F}\) and we wish to lower the potential difference across the plates by \(60.0 \mathrm{V},\) what magnitude of charge will we have to remove from each plate?
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A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) and is connected to a \(12-\mathrm{V}\) battery. (a) What is the magnitude of the charge on each plate? (b) If the plate separation is doubled while the plates remain connected to the battery, what happens to the charge on each plate and the electric field between the plates?
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