An electron is to be accelerated from 3.00×1063.00\$$times10^6 m/s to 8.00×1068.00$$\$$times10^6 m/s. $$Through what potential difference must the electron pass to accomplish this?

Ch 23: Electric Potential

Young & Freedman Calc - University Physics 15th Edition

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Young & Freedman Calc 15th Edition

Ch 23: Electric Potential

Problem 20a

Chapter 23, Problem 20a

An electron is to be accelerated from $3.00 \times 10^{6}$ m/s to $8.00 \times 10^{6}$ m/s. Through what potential difference must the electron pass to accomplish this?

Verified step by step guidance

First, understand that the problem involves the kinetic energy change of an electron due to acceleration. The kinetic energy (KE) of an electron can be calculated using the formula:

K_{e} = \frac{1}{2} m v^{2}

, where

m

is the mass of the electron and

v

is its velocity.

Calculate the initial kinetic energy using the initial velocity

v_{i}

= 3.00x10^6 m/s. Substitute this value into the kinetic energy formula:

K_{i} = \frac{1}{2} m {v_{i}}^{2}

Calculate the final kinetic energy using the final velocity

v_{f}

= 8.00x10^6 m/s. Substitute this value into the kinetic energy formula:

K_{f} = \frac{1}{2} m {v_{f}}^{2}

Determine the change in kinetic energy, which is the difference between the final and initial kinetic energies:

ΔK = K_{f} - K_{i}

Relate the change in kinetic energy to the potential difference using the work-energy principle. The work done on the electron is equal to the change in kinetic energy, and this work is also equal to the charge of the electron multiplied by the potential difference:

e V = ΔK

. Solve for the potential difference

V

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion, calculated using the formula KE = 0.5 * m * v^2, where m is mass and v is velocity. In this problem, the change in kinetic energy of the electron as it accelerates from one velocity to another is crucial for determining the potential difference required.

Electric Potential Difference

Electric potential difference, measured in volts, is the work done per unit charge to move a charge between two points in an electric field. It relates to the energy change of a charged particle, such as an electron, as it moves through an electric field, and is essential for calculating the energy required to change the electron's velocity.

Conservation of Energy

The conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another. In this context, the work done on the electron by the electric field (potential energy) is converted into kinetic energy, allowing us to equate the change in kinetic energy to the work done by the potential difference.

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Related Practice

Textbook Question

Two point charges $q_1 = +2.40$ nC and $q_2 = -6.50$ nC are $0.100$ m apart. Point $A$ is midway between them; point $B$ is $0.080$ m from $q_1$ and $0.060$ m from $q_2$ (Fig. E $23.19$ ). Take the electric potential to be zero at infinity. Find the potential at point $B$ .

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Textbook Question

Two point charges of equal magnitude $Q$ are held a distance $d$ apart. Consider only points on the line passing through both charges. If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?).

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Textbook Question

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are $4.98$ V and $16.2$ V/m, respectively. (Take $V = 0$ at infinity.) What is the distance to the point charge?

Textbook Question

Two point charges $q sub 1 equals positive 2.40$ nC and $q_{2} = - 6.50$ nC are $0.100$ m apart. Point $A$ is midway between them; point $B$ is $0.080$ m from $q sub 1$ and $0.060$ m from $q sub 2$ (Fig. E $23.19$ ). Take the electric potential to be zero at infinity. Find the potential at point $A$ .

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Textbook Question

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are $4.98$ V and $16.2$ V/m, respectively. (Take $V equals 0$ at infinity.) What is the magnitude of the charge?

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Textbook Question

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are $4.98$ V and $16.2$ V/m, respectively. (Take $V = 0$ at infinity.) Is the electric field directed toward or away from the point charge?

An electron is to be accelerated from 3.00×1063.00\(\times\)10^6 m/s to 8.00×1068.00\(\times\)10^6 m/s. Through what potential difference must the electron pass to accomplish this?

Verified video answer for a similar problem:

Key Concepts

Kinetic Energy

Electric Potential Difference

Conservation of Energy

An electron is to be accelerated from $3.00 \times 10^{6}$ m/s to $8.00 \times 10^{6}$ m/s. Through what potential difference must the electron pass to accomplish this?