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Chapter 23: Problem 45

A \(2.50-\mathrm{mg}\) dust particle with a charge of \(1.00 \mu \mathrm{C}\) falls at a point \(x=2.00 \mathrm{~m}\) in a region where the electric potential varies according to \(V(x)=\left(2.00 \mathrm{~V} / \mathrm{m}^{2}\right) x^{2}-\left(3.00 \mathrm{~V} / \mathrm{m}^{3}\right) x^{3}\) With what acceleration will the particle start moving after it touches down?

Short Answer

Expert verified
Question: Calculate the acceleration of a dust particle with a charge of 1.00 µC and mass 2.50 mg when it falls in a region with varying electric potential \(V(x) = (2.00 \mathrm{V/m^2})x^2 - (3.00 \mathrm{V/m^3})x^3\) at position x = 2.00 m. Answer: To find the acceleration of the dust particle, follow these steps: 1. Determine the electric field at the given position x: \(E(x) = -(4.00 \mathrm{V/m^2})(2.00 \mathrm{m}) + (9.00 \mathrm{V/m^3})(2.00 \mathrm{m})^2\) 2. Calculate the electric force on the dust particle: \(F = (1.00 \times 10^{-6}\mathrm{C})((-4.00 \mathrm{V/m^2})(2.00 \mathrm{m}) + (9.00 \mathrm{V/m^3})(2.00 \mathrm{m})^2)\) 3. Calculate the acceleration of the particle: \(a = \frac{(1.00 \times 10^{-6} \mathrm{C})((-4.00 \mathrm{V/m^2})(2.00 \mathrm{m}) + (9.00 \mathrm{V/m^3})(2.00 \mathrm{m})^2)}{(2.50 \times 10^{-6} \mathrm{kg})}\) Compute the final value of acceleration to find the answer.

Step by step solution

01

Determine the electric field at given position x

First, we need to know the electric field at position x. To find that, we'll take the derivative of the electric potential function V(x) with respect to x. The electric field, E, can be found by taking the negative of the derivative, or E = -dV/dx. Given function for electric potential: \(V(x) = (2.00 \mathrm{V/m^2})x^2 - (3.00 \mathrm{V/m^3})x^3\) Now, take the derivative of V(x) with respect to x: \(dV/dx = (4.00 \mathrm{V/m^2})x - (9.00 \mathrm{V/m^3})x^2\) Finally, find the electric field at the given position by taking the negative of dV/dx: \(E(x) = -(4.00 \mathrm{V/m^2})(2.00 \mathrm{m}) + (9.00 \mathrm{V/m^3})(2.00 \mathrm{m})^2\)
02

Calculate the electric force on the dust particle

Now that we know the electric field at position x, we can find the electric force acting on the dust particle using Coulomb's law: \(F = qE\) Plug in the given charge of the dust particle (q = 1.00 µC = \(1.00 \times 10^{-6}\mathrm{C}\)) and the calculated electric field at position x: \(F = (1.00 \times 10^{-6}\mathrm{C})((-4.00 \mathrm{V/m^2})(2.00 \mathrm{m}) + (9.00 \mathrm{V/m^3})(2.00 \mathrm{m})^2)\)
03

Calculate the acceleration of the particle

Now, we can find the acceleration of the dust particle using Newton's second law: \(a = \frac{F}{m}\) The mass of the particle is given as m = \(2.50\,\mathrm{mg} = 2.50 \times 10^{-6}\,\mathrm{kg}\); use the mass and the calculated electric force to find the acceleration: \(a = \frac{(1.00 \times 10^{-6} \mathrm{C})((-4.00 \mathrm{V/m^2})(2.00 \mathrm{m}) + (9.00 \mathrm{V/m^3})(2.00 \mathrm{m})^2)}{(2.50 \times 10^{-6} \mathrm{kg})}\) Compute the final value of acceleration to complete the solution.

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Key Concepts

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

Coulomb's Law
Coulomb's Law is essential for understanding the interaction between charged particles. It states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:
\[ F = k \frac{q_1 q_2}{r^2} \]
Where:
  • \( F \) is the magnitude of the electric force,
  • \( k \) is Coulomb's constant (\( 8.9875 \times 10^9 \mathrm{N m^2/C^2} \)),
  • \( q_1 \) and \( q_2 \) are the charges,
  • \( r \) is the distance between the charges.
In the exercise, Coulomb's Law helps us understand how the charge of the dust particle interacts with the electric field, resulting in an electric force.
Electric Force
The electric force plays a pivotal role in defining how charges influence each other. In the context of the exercise, once we have determined the electric field at a point, we can calculate the force experienced by a charge in that field using the formula:
\[ F = qE \]
Where:
  • \( F \) is the electric force,
  • \( q \) is the charge of the particle, and
  • \( E \) is the electric field strength at that point.
By applying Coulomb's Law, we can see that the electric field is essentially the force per unit charge. For the dust particle in the problem, we apply the given charge to the calculated electric field to find the force it experiences.
Electric Field
The electric field concept is central to understanding the behavior of charged particles in space. It's a vector field that surrounds electric charges and is defined as the electric force per unit charge. The electric field at any point in space can be calculated by the formula:
\[ E = \frac{F}{q} \]
Where \( F \) is the force exerted on a small positive test charge and \( q \) is the magnitude of the test charge. The direction of the electric field is the direction of the force it would exert on a positive test charge. A core aspect of this exercise is deriving the electric field from the given electric potential, which involves taking the negative gradient (or derivative, in one dimension) of the potential function: \( E = -\frac{dV}{dx} \).
Derivative of Electric Potential
To determine an electric field from an electric potential, we use the concept of the derivative of the electric potential. The derivative gives us the rate at which the potential changes with respect to position, and when we take the negative of this derivative, we get the electric field. This is because, intuitively, a charge tends to move from higher to lower potential, and thus the 'negative slope' gives us the direction of the field. The exercise provided demonstrates this with:
\[ E(x) = -\frac{dV}{dx} \]
Upon calculation, we plug the position \( x \) into the derivative to calculate the electric field at that specific point, which affects the dust particle. Henceforth, the exercise's guidance to take the derivative of the electric potential function is critical.
Newton's Second Law
Newton's Second Law is fundamental in physics, relating the motion of an object to the forces acting on it. Stated simply, the law can be expressed as:
\[ F = ma \]
  • Where \( F \) is the total force exerted on an object,
  • \( m \) is the mass of the object, and
  • \( a \) is the acceleration of the object.
In the context of our exercise, this law allows us to calculate the acceleration of the dust particle once we have determined the electric force acting on it. By rearranging the formula to solve for acceleration (\( a = F/m \)), we find how quickly the particle will begin to move when influenced by the electric force we previously calculated.

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Most popular questions from this chapter

A charge of \(0.681 \mathrm{nC}\) is placed at \(x=0 .\) Another charge of \(0.167 \mathrm{nC}\) is placed at \(x_{1}=10.9 \mathrm{~cm}\) on the \(x\) -axis. a) What is the combined electrostatic potential of these two charges at \(x=20.1 \mathrm{~cm},\) also on the \(x\) -axis? b) At which point(s) on the \(x\) -axis does this potential have a minimum?

The electric potential energy of a continuous charge distribution can be found in a way similar to that used for systems of point charges in Section \(23.6,\) by breaking the distribution up into suitable pieces. Find the electric potential energy of an arbitrary spherically symmetrical charge distribution, \(\rho(r) .\) Do not assume that \(\rho(r)\) represents a point charge, that it is constant, that it is piecewise-constant, or that it does or does not end at any finite radius, \(r\). Your expression must cover all possibilities. Your expression may include an integral or integrals that cannot be evaluated without knowing the specific form of \(\rho(r) .\) (Hint: A spherical pearl is built up of thin layers of nacre added one by one.)

In which situation is the electric potential the highest? a) at a point \(1 \mathrm{~m}\) from a point charge of \(1 \mathrm{C}\) b) at a point \(1 \mathrm{~m}\) from the center of a uniformly charged spherical shell of radius \(0.5 \mathrm{~m}\) with a total charge of \(1 \mathrm{C}\) c) at a point \(1 \mathrm{~m}\) from the center of a uniformly charged rod of length \(1 \mathrm{~m}\) and with a total charge of \(1 \mathrm{C}\) d) at a point \(2 \mathrm{~m}\) from a point charge of \(2 \mathrm{C}\) e) at a point \(0.5 \mathrm{~m}\) from a point charge of \(0.5 \mathrm{C}\)

A Van de Graaff generator has a spherical conductor with a radius of \(25.0 \mathrm{~cm}\). It can produce a maximum electric field of \(2.00 \cdot 10^{6} \mathrm{~V} / \mathrm{m}\). What are the maximum voltage and charge that it can hold?

A point charge of \(+2.0 \mu C\) is located at \((2.5 \mathrm{~m}, 3.2 \mathrm{~m})\) A second point charge of \(-3.1 \mu \mathrm{C}\) is located at \((-2.1 \mathrm{~m}, 1.0 \mathrm{~m})\). a) What is the electric potential at the origin? b) Along a line passing through both point charges, at what point(s) is (are) the electric potential(s) equal to zero?
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