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Chapter 21: Problem 64

A small sphere with positive charge \(q\) and mass \(m\) is released from rest in a uniform electric field \(\vec{E}\) that is directed vertically upward. The magnitude of the field is large enough for the sphere to travel upward when it is released. How long does it take the sphere to travel upward a distance \(d\) after being released from rest? Give your answer in terms of \(q, m, d, E,\) and the acceleration due to gravity, \(g\).

Short Answer

Expert verified
The amount of the time it will take the sphere to travel upward distance \(d\) after being released from rest can be expressed as \(t = \sqrt{ \frac{2d}{((qE - mg) / m)} }= \sqrt{ \frac{2md}{qE - mg} }\)

Step by step solution

01

Identify the forces acting on the sphere

The forces acting on the sphere are the gravitational force and the electric force. The gravitational force can be represented as \(F_g=mg\) where \(g\) is the acceleration due to gravity. The electric force can be represented as \(F_e=qE\) where \(E\) is the electric field and \(q\) is the charge of the sphere.
02

Understand the resultant force

The resultant force will be the force due to the electric field minus the force due to gravity because these two forces act in opposing directions. This can be represented as \(F=F_e - F_g\). Substituting the expressions from Step 1 results in \(F=qE-mg\).
03

Apply Newton's second law

Newton's second law states that the force acting on an object is equal to its mass times its acceleration. This can be represented as \(F=ma\). Setting the expressions from Step 2 and Step 3 equal to each other gives \(qE-mg=ma\). Solving for \(a\) gives us \(a = (qE - mg) / m\).
04

Calculate time

The equation for motion with constant acceleration is \(d = ut + 0.5*a*t^2\) where \(d\) is the distance, \(u\) is initial velocity and \(a\) is acceleration. Here, \(u\) being \(0\) (the ball is initially at rest) simplifies the equation to \(d = 0.5*a*t^2\). Solving for \(t\) gives us \(t = \sqrt{(2d)/a}\). Subsituting the expression for \(a\) and simplifying will give the answer.

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Key Concepts

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

Electric Force
The concept of electric force is integral to understanding how charged objects interact with an electric field. When a charged particle, such as a small sphere with a positive charge \(q\), is placed in a uniform electric field \(\vec{E}\), it experiences an electric force. This force can be calculated using the equation \(F_e = qE\), where \(E\) represents the magnitude of the electric field.

This force acts in the direction of the electric field for a positively charged object, and opposite to the field for a negatively charged object. In our exercise, as the field is directed vertically upward and the charge is positive, the electric force propels the sphere upward.
Gravitational Force
The gravitational force is a fundamental force that acts between any two masses in the universe. For objects near the surface of the Earth, this force causes them to experience an acceleration known as the acceleration due to gravity, \(g\), which has a magnitude of approximately \(9.81 \text{m/s}^2\) downward.

For a small sphere with mass \(m\), the gravitational force can be expressed as \(F_g = mg\). This force acts downward, opposite to the electric force in our scenario. Understanding this force is essential to determine the net force acting on the sphere and, consequently, its motion.
Newton's Second Law
Newton's second law of motion is foundational in physics and states that the force \(F\) acting on an object is equal to the mass \(m\) of the object multiplied by its acceleration \(a\), expressed as \(F=ma\).

In the context of our exercise, Newton's second law allows us to calculate the acceleration of the sphere by equating the net force acting on the sphere to the product of its mass and acceleration. When the electric force \(F_e\) and the gravitational force \(F_g\) are combined, we can find the resultant force, and use it to determine the acceleration of the sphere.
Constant Acceleration Motion
Constant acceleration motion refers to the motion of an object when it is subjected to a constant net force, resulting in a constant acceleration. The equations of motion for constant acceleration are invaluable tools in kinematics for predicting the future position or velocity of an object.

In our case, with the sphere starting from rest and being propelled by the net force from electric and gravitational fields, we have a scenario of constant acceleration. The equation \(d = ut + 0.5at^2\) expresses the distance \(d\) traveled over time \(t\), where \(u\) is the initial velocity and \(a\) is acceleration. For the sphere released from rest, \(u\) is zero and we can solve for the time \(t\) taken to travel a distance \(d\) using the sphere鈥檚 constant acceleration.

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Most popular questions from this chapter

Inkjet printers can be described as either continuous or drop-on-demand. In a continuous inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. You are part of an engineering group working on the design of such a printer. Each ink drop will have a mass of \(1.4 \times 10^{-8} \mathrm{~g}\). The drops will leave the nozzle and travel toward the paper at \(50 \mathrm{~m} / \mathrm{s}\) in a horizontal direction, passing through a charging unit that gives each drop a positive charge \(q\) by removing some electrons from it. The drops will then pass between parallel deflecting plates, \(2.0 \mathrm{~cm}\) long, where there is a uniform vertical electric field with magnitude \(8.0 \times 10^{4} \mathrm{~N} / \mathrm{C}\). Your team is working on the design of the charging unit that places the charge on the drops. (a) If a drop is to be deflected \(0.30 \mathrm{~mm}\) by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? How many electrons must be removed from the drop to give it this charge? (b) If the unit that produces the stream of drops is redesigned so that it produces drops with a speed of \(25 \mathrm{~m} / \mathrm{s},\) what \(q\) value is needed to achieve the same \(0.30 \mathrm{~mm}\) deflection?

A point charge is placed at each corner of a square with side length a. All charges have magnitude \(q\). Two of the charges are positive and two are negative (Fig. E21.38). What is the direction of the net electric field at the center of the square due to the four charges, and what is its magnitude in terms of \(q\) and \(a\) ?

Two small aluminum spheres, each having mass \(0.0250 \mathrm{~kg}\), are separated by \(80.0 \mathrm{~cm}\). (a) How many electrons does each sphere contain? (The atomic mass of aluminum is \(26.982 \mathrm{~g} / \mathrm{mol}\), and its atomic number is \(13 .\) ) (b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude \(1.00 \times 10^{4} \mathrm{~N}\) (roughly 1 ton)? Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?

Four identical charges \(Q\) are placed at the corners of a square of side \(L\). (a) In a free-body diagram, show all of the forces that act on one of the charges. (b) Find the magnitude and direction of the total force exerted on one charge by the other three charges.

A point charge is at the origin. With this point charge as the source point, what is the unit vector \(\hat{r}\) in the direction of the field point (a) at \(x=0, y=-1.35 \mathrm{~m}\) (b) at \(x=12.0 \mathrm{~cm}, y=12.0 \mathrm{~cm} ;\) (c) at \(x=-1.10 \mathrm{~m}, y=2.60 \mathrm{~m} ?\) Express your results in terms of the unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\)
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