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There is an electric field of magnitude \(150 .\) N/C, directed vertically downward, near the surface of the Earth. Find the acceleration (magnitude and direction) of an electron released near the Earth's surface.

Short Answer

Expert verified
Answer: The acceleration of the electron is \(2.63 \times 10^{13} \text{ m/s}^2\) directed vertically upward.

Step by step solution

01

Identify the given information

We are given an electric field with magnitude 150 N/C directed vertically downward. We also know the charge of an electron is \(-1.6 \times 10^{-19}\) C and the mass of an electron is \(9.11 \times 10^{-31}\) kg.
02

Calculate the force acting on the electron

Using the formula F = qE, we can calculate the force acting on the electron. We have to take the negative sign of electron charge into account while calculating the force. F = qE = \((-1.6 \times 10^{-19} \text{ C})(150 \text{ N/C}) = -2.4 \times 10^{-17}\) N
03

Determine the direction of the force

Since the force is negative and the electric field is directed vertically downward, the force on the electron is directed vertically upward.
04

Calculate the acceleration of the electron

Using Newton's second law, F = ma, we can calculate the acceleration of the electron: a = F/m = \((-2.4 \times 10^{-17} \text{ N})/(9.11 \times 10^{-31} \text{ kg}) = 2.63 \times 10^{13} \text{ m/s}^2\)
05

Determine the direction of the acceleration

Since the force acting on the electron is vertically upward, the acceleration of the electron will also be vertically upward.
06

Write the final answer

The acceleration of the electron is \(2.63 \times 10^{13} \text{ m/s}^2\) directed vertically upward.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field Strength
Understanding the concept of electric field strength is fundamental when studying how charges interact with fields. In essence, an electric field represents the force per unit charge exerted on a small positive test charge placed within it. Electric field strength is measured in newtons per coulomb (N/C) and is a vector quantity, meaning it has both magnitude and direction.

For an electron moving through such a field, the interaction between its own negative charge and the field's character determines the force exerted on the electron. When the field strength is declared as 150 N/C as in our exercise, this implicitly expresses not only the potency of the field but also sets the stage for calculating the resulting motion — or acceleration — of charged particles within it.
Electron Charge
The charge of an electron is a fundamental constant in physics and is essential for understanding electric field interactions. Denoted by a negative sign, the electron's charge has a magnitude of approximately equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to equal to nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnannnnnnnnnnnnnnnnnnnnnnnn Souls-19C. This opposes the positive test charge used to conceptualize fields and injects interesting dynamics when an electron is subjected to an electric field, such as in our example.

In the context of the problem, this relatively tiny charge results in a tangible force when multiplied by the electric field's strength. Here, knowing the electron's charge is of utmost importance as it is directly proportional to the force exerted by the electric field.
Newton's Second Law
Newton's second law is perhaps one of the most well-known physics principles and is often summarized by the equation F=ma, which tells us that force equals mass times acceleration. It's the backbone of classical mechanics and describes how an object will accelerate when subjected to a net external force.

For our electron scenario, this law translates to the principle that governs the determination of its acceleration. Since the electron's mass is a known quantity, and we've previously calculated the force acting on it due to the electric field, Newton's second law allows us to solve for the electron's acceleration directly. Newton's law helps us understand that the direction of the acceleration will be opposite to the direction of the force when considering the negative charge of the electron.
Acceleration Calculation
The calculation of acceleration is a straightforward application of Newton's second law, once the force acting on the object is known. Acceleration can be defined as the rate of change of velocity over time and is expressed in units of meters per second squared (m/s^2).

For calculating an electron's acceleration in an electric field, we divide the net force acting on it by its mass. As illustrated in the problem, the direction of the acceleration is determined by the direction of the force acting on the electron. Despite the electron's incredibly small mass, the vast magnitude of acceleration resulting from this division underscores just how sensitive electrons are to electrical forces. This sensitivity is why even small electric fields can have significant effects on electron behavior.

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Most popular questions from this chapter

At which of the following locations is the electric field the strongest? a) a point \(1 \mathrm{~m}\) from a \(1 \mathrm{C}\) point charge b) a point \(1 \mathrm{~m}\) (perpendicular distance) from the center of a \(1-\mathrm{m}\) -long wire with \(1 \mathrm{C}\) of charge distributed on it c) a point \(1 \mathrm{~m}\) (perpendicular distance) from the center of a \(1-\mathrm{m}^{2}\) sheet of charge with \(1 \mathrm{C}\) of charge distributed on it d) a point \(1 \mathrm{~m}\) from the surface of a charged spherical shell of charge \(1 \mathrm{C}\) with a radius of \(1 \mathrm{~m}\) e) a point \(1 \mathrm{~m}\) from the surface of a charged spherical shell of charge \(1 \mathrm{C}\) with a radius of \(0.5 \mathrm{~m}\)

\( \mathrm{~A}+48.00-\mathrm{nC}\) point charge is placed on the \(x\) -axis at \(x=4.000 \mathrm{~m},\) and \(\mathrm{a}-24.00-\mathrm{n} \mathrm{C}\) point charge is placed on the \(y\) -axis at \(y=-6.000 \mathrm{~m} .\) What is the direction of the electric field at the origin?

A uniformly charged rod of length \(L\) with total charge \(Q\) lies along the \(y\) -axis, from \(y=0\) to \(y=L\). Find an expression for the electric field at the point \((d, 0)\) (that is, the point at \(x=d\) on the \(x\) -axis).

Consider an electric dipole on the \(x\) -axis and centered at the origin. At a distance \(h\) along the positive \(x\) -axis, the magnitude of electric field due to the electric dipole is given by \(k(2 q d) / h^{3} .\) Find a distance perpendicular to the \(x\) axis and measured from the origin at which the magnitude of the electric field stays the same.

A long conducting wire with charge distribution \(\lambda\) and radius \(r\) produces an electric field of \(2.73 \mathrm{~N} / \mathrm{C}\) just outside the surface of the wire. What is the magnitude of the electric field just outside the surface of another wire with charge distribution \(0.81 \lambda\) and radius \(6.5 r ?\)

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