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Chapter 11: Problem 1588

If an electron moves from rest from a point at which potential is 50 volt, to another point at which potential is 70 volt, then its kinetic energy in the final state will be \(\ldots .\) (A) \(1 \mathrm{~N}\) (B) \(3.2 \times 10^{-18} \mathrm{~J}\) (C) \(3.2 \times 10^{-10} \mathrm{~J}\) (D) 1 dyne

Short Answer

Expert verified
The short answer based on the provided step-by-step solution is: The final kinetic energy of the electron after moving from an initial potential of 50 V to a final potential of 70 V is \(3.2 \times 10^{-18} \mathrm{~J}\) (Option B).

Step by step solution

01

Understand the concepts of potential and kinetic energy

Potential energy is the energy an object possesses because of its position in a force field or that a system has stored because of the relative positions of its components. In the case of an electron, its potential energy is due to its position in an electric field. Kinetic energy is the energy an object possesses because of its motion. In this exercise, we will find the change in kinetic energy as the electron moves from one potential to another.
02

Find the change in potential energy

The electron moves from an initial potential of 50 V to a final potential of 70 V. The change in potential energy, ∆U, can be found by the equation: ∆U = q * ∆V Where q is the charge of an electron (q = -1.6 × 10^(-19) C), and ∆V is the change in potential, which in this case is (70 V - 50 V).
03

Calculate the change in potential energy

Calculate the change in potential energy using the equation from step 2: ∆U = (-1.6 × 10^(-19) C) * (70 V - 50 V) ∆U = -1.6 × 10^(-19) C * (20 V) ∆U = -3.2 × 10^(-18) J Since the electron is moving from a lower potential (50 V) to higher potential (70 V), it's losing potential energy, thus the negative sign.
04

Determine the change in kinetic energy

By the conservation of energy principle, the change in potential energy is equal to the change in kinetic energy: ∆K = -∆U ∆K = 3.2 × 10^(-18) J
05

Identify the correct answer

Now we can see that the final kinetic energy of the electron after moving from 50 V to 70 V is 3.2 × 10^(-18) J. Comparing this value with the given options, we find that the correct answer is: (B) \(3.2 \times 10^{-18} \mathrm{~J}\)

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Most popular questions from this chapter

Two point charges \(-q\) and \(+q\) are located at points \((0,0,-a)\) and $(0,0, a)\( respectively. The potential at a point \)(0,0, z)\( where \)z>a\( is \)\ldots \ldots$ (A) $\left[(2 \mathrm{q} a) /\left\\{4 \pi \epsilon_{0}\left(z^{2}+a^{2}\right)\right\\}\right]$ (B) \(\left[\mathrm{q} /\left(4 \pi \epsilon_{0} \mathrm{a}\right)\right]\) (C) \(\left[\right.\) (qa) \(\left./\left(4 \pi \in_{0} z^{2}\right)\right]\) (D) $\left[(2 q a) /\left\\{4 \pi \epsilon_{0}\left(z^{2}-a^{2}\right)\right\\}\right]$

A point charge \(q\) is situated at a distance \(r\) from one end of a thin conducting rod of length \(\mathrm{L}\) having a charge \(\mathrm{Q}\) (uniformly distributed along its length). The magnitude of electric force between the two, is ...... \((\mathrm{A})[(2 \mathrm{kqQ}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\) (B) \([(\mathrm{kq} \mathrm{Q}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\) (C) \([(\mathrm{kqQ}) / \mathrm{r}(\mathrm{r}-\mathrm{L})]\) (D) \([(\mathrm{kQ}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\)

A parallel plate capacitor of capacitance \(5 \mu \mathrm{F}\) and plate separation \(6 \mathrm{~cm}\) is connected to a \(1 \mathrm{~V}\) battery and charged. A dielectric of dielectric constant 4 and thickness \(4 \mathrm{~cm}\) is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is (A) \(2 \mu \mathrm{c}\) (B) \(5 \mu \mathrm{c}\) (C) \(3 \mu \mathrm{c}\) (D) \(10 \mu \mathrm{c}\)

A simple pendulum of period \(\mathrm{T}\) has a metal bob which is negatively charged. If it is allowed to oscillate above a positively charged metal plate, its period will ...... (A) Remains equal to \(\mathrm{T}\) (B) Less than \(\mathrm{T}\) (C) Infinite (D) Greater than \(\mathrm{T}\)

The electric potential \(\mathrm{V}\) is given as a function of distance \(\mathrm{x}\) (meter) by $\mathrm{V}=\left(5 \mathrm{x}^{2}+10 \mathrm{x}-9\right)\( volt. Value of electric field at \)\mathrm{x}=1$ is \(\ldots \ldots\) \((\mathrm{A})-20(\mathrm{v} / \mathrm{m})\) (B) \(6(\mathrm{v} / \mathrm{m})\) (C) \(11(\mathrm{v} / \mathrm{m})\) (D) \(-23(\mathrm{v} / \mathrm{m})\)
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