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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 of \(+16 \mathrm{c}\) and \(-9 \mathrm{c}\) are placed $8 \mathrm{~cm}\( apart in air \)\ldots \ldots\(.. distance of a point from \)-9$ c charge at which the resultant electric field is zero. (A) \(24 \mathrm{~cm}\) (B) \(9 \mathrm{~cm}\) (C) \(16 \mathrm{~cm}\) (D) \(35 \mathrm{~cm}\)

A long string with a charge of \(\lambda\) per unit length passes through an imaginary cube of edge \(\ell\). The maximum possible flux of the electric field through the cube will be ....... (A) \(\sqrt{3}\left(\lambda \ell / \in_{0}\right)\) (B) \(\left(\lambda \ell / \in_{0}\right)\) (C) \(\sqrt{2}\left(\lambda \ell / \in_{0}\right)\) (D) \(\left[\left(6 \lambda \ell^{2}\right) / \epsilon_{0}\right]\)

Point charges \(4 \mu \mathrm{c}\) and \(2 \mu \mathrm{c}\) are placed at the vertices \(\mathrm{P}\) and Q of a right angle triangle \(P Q R\) respectively. \(Q\) is the right angle, \(\mathrm{PR}=2 \times 10^{-2} \mathrm{~m}\) and \(\mathrm{QR}=10^{-2} \mathrm{~m}\). The magnitude and direction of the resultant electric field at \(\mathrm{R}\) is \(\ldots \ldots\) (A) \(4.28 \times 10^{9} \mathrm{NC}^{-1}, 45^{\circ}\) (B) \(2.38 \times 10^{8} \mathrm{NC}^{-1}, 40.9^{\circ}\) (C) \(1.73 \times 10^{4} \mathrm{NC}^{-1}, 34.7^{\circ}\) (D) \(4.9 \times 10^{10} \mathrm{NC}^{-1}, 34.7^{\circ}\)

A simple pendulum consists of a small sphere of mass \(\mathrm{m}\) suspended by a thread of length \(\ell\). The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength \(\mathrm{E}\) directed Vertically upwards. If the electrostatic force acting on the sphere is less than gravitational force the period of pendulum is (A) $\mathrm{T}=2 \pi[\ell /\\{\mathrm{g}-(\mathrm{q} \mathrm{E} / \mathrm{m})\\}]^{(1 / 2)}$ (B) \(\mathrm{T}=2 \pi(\ell / \mathrm{g})^{(1 / 2)}\) \(\left.\left.\left.\mathrm{m}_{\mathrm{}}\right\\}\right\\}\right]^{(1 / 2)}\) (D) \(\mathrm{T}=2 \pi[(\mathrm{m} \ell / \mathrm{qE})]^{(1 / 2)}\) (C) \(\mathrm{T}=2 \pi[\ell /\\{\mathrm{g}+(\mathrm{qE} / \mathrm{t}\)

Three charges \(2 q,-q,-q\) are located at the vertices of an equilateral triangle. At the centre of the triangle. (A) The Field is Zero but Potential is non - zero (B) The Field is non - Zero but Potential is zero (C) Both field and Potential are Zero (D) Both field and Potential are non - Zero
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