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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

An electric dipole with a dipole moment of magnitude \(p\) is placed at various orientations in an electric field \(\vec{E}\) that is directed to the left. (a) What orientation of the dipole will result in maximum torque directed into the page? What then is the electric potential energy? (b) What orientation of the dipole will give zero torque and maximum electric potential energy? What type of equilibrium is this: stable, unstable, or neutral?

A uniform line of charge with length \(20.0 \mathrm{~cm}\) is along the \(x\) -axis, with its midpoint at \(x=0 .\) Its charge per length is \(+4.80 \mathrm{nC} / \mathrm{m}\) A small sphere with charge \(-2.00 \mu \mathrm{C}\) is located at \(x=0, y=5.00 \mathrm{~cm}\) What are the magnitude and direction of the force that the charged sphere exerts on the line of charge?

Three parallel sheets of charge, large enough to be treated as infinite sheets, are perpendicular to the \(x\) -axis. Sheet \(A\) has surface charge density \(\sigma_{A}=+8.00 \mathrm{nC} / \mathrm{m}^{2}\). Sheet \(B\) is \(4.00 \mathrm{~cm}\) to the right of sheet \(A\) and has surface charge density \(\sigma_{B}=-4.00 \mathrm{nC} / \mathrm{m}^{2} .\) Sheet \(C\) is \(4.00 \mathrm{~cm}\) to the right of sheet \(B,\) so is \(8.00 \mathrm{~cm}\) to the right of sheet \(A,\) and has surface charge density \(\sigma_{C}=+6.00 \mathrm{nC} / \mathrm{m}^{2}\). What are the magnitude and direction of the resultant electric field at a point that is midway between sheets \(B\) and \(C,\) or \(2.00 \mathrm{~cm}\) from each of these two sheets?

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.

Two tiny spheres of mass \(6.80 \mathrm{mg}\) carry charges of equal magnitude, \(72.0 \mathrm{nC}\), but opposite sign. They are tied to the same ceiling hook by light strings of length \(0.530 \mathrm{~m}\). When a horizontal uniform electric field \(E\) that is directed to the left is turned on, the spheres hang at rest with the angle \(\theta\) between the strings equal to \(58.0^{\circ}\) (Fig. P21.74). (a) Which ball (the one on the right or the one on the left) has positive charge? (b) What is the magnitude \(E\) of the field?
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