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A charge \(q\) in a uniform electric field \(\mathbf{E}_{0}\) experiences a constant force \(\mathbf{F}=q \mathbf{E}_{0}\). (a) Show that this force is conservative and verify that the potential energy of the charge at position \(\mathbf{r}\) is \(U(\mathbf{r})=-q \mathbf{E}_{0} \cdot \mathbf{r}\). (b) By doing the necessary derivatives, check that \(\mathbf{F}=-\nabla U\).

Short Answer

Expert verified
Force is conservative. Potential: \(U(\mathbf{r}) = -q \mathbf{E}_{0} \cdot \mathbf{r}\), \(\mathbf{F} = -\nabla U\).

Step by step solution

01

Understand the Problem

We need to demonstrate that the force \( \mathbf{F} = q \mathbf{E}_{0} \) is conservative and show that the potential energy is \( U(\mathbf{r}) = -q \mathbf{E}_{0} \cdot \mathbf{r} \). Additionally, we should verify the relationship \( \mathbf{F} = -abla U \).
02

Define a Conservative Force

A force is considered conservative if the work done by the force around any closed path is zero. Equivalently, a force is conservative if there exists a scalar potential energy function \( U \) such that the force can be written as \( \mathbf{F} = -abla U \).
03

Propose the Potential Energy Function

Assume the potential energy function \( U(\mathbf{r}) = -q \mathbf{E}_{0} \cdot \mathbf{r} \). This form is suggested because it represents the standard approach to define potential energy in a uniform electric field.
04

Evaluate the Gradient of U

Calculate the gradient \( abla U \) where \( U(\mathbf{r}) = -q \mathbf{E}_{0} \cdot \mathbf{r} \). The gradient operator is applied to each component of \( \mathbf{r} \), and \( \mathbf{E}_{0} \) being constant results in \( abla U = -q \mathbf{E}_{0} \).
05

Check the Force from Gradient

From the relationship \( \mathbf{F} = -abla U \), substitute for \( abla U \) giving \( \mathbf{F} = q \mathbf{E}_{0} \). This is consistent with the given \( \mathbf{F} = q \mathbf{E}_{0} \), confirming the force as conservative.
06

Verify the Conservative Condition

As \( \mathbf{F} = -abla U \) holds true, it implies that \( \mathbf{F} \) is a conservative force. Thus, the work done in a closed path in a uniform electric field is zero, consistent with the properties of conservative forces.

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Key Concepts

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

Electric Potential Energy
Electric potential energy is a concept that helps us understand how charged particles interact with electric fields. When a charge is placed in an electric field, it experiences a force which can do work on the charge. This interaction is quantified by the electric potential energy.

The electric potential energy of a charge in a uniform electric field is given by the expression:
  • \( U(\mathbf{r}) = -q \mathbf{E}_{0} \cdot \mathbf{r} \)
Here, \( q \) is the charge, \( \mathbf{E}_{0} \) is the uniform electric field, and \( \mathbf{r} \) is the position vector of the charge with respect to a reference point. The negative sign indicates that the electric potential energy decreases when the charge moves in the direction of the field.

This formula allows us to calculate the potential energy for any position of the charge within the field. Understanding this helps to predict how the charge will behave as it moves through the electric field.
Uniform Electric Field
A uniform electric field is one where the electric field strength and direction are constant at every point in the space between two parallel plates or any situation designed to create such a field. This consistency makes it relatively easy to work with mathematically and practically.

In the case of a uniform electric field:
  • The field lines are parallel and equally spaced.
  • For any charge entering the field, the force experienced remains constant, given by \( \mathbf{F}=q \mathbf{E}_{0} \).
  • This force acts in the direction of the electric field if the charge is positive, and opposite if the charge is negative.
Electric fields are often represented by vectors in physics problems, where the vector’s length corresponds to the field's strength. In a uniform field, the implications for forces and movements are simple, because their behavior is quite predictable. A great way to visualize it is as a flat, even surface pushing on all charges equally.
Gradient Operator
The gradient operator, denoted by \( abla \), is a key mathematical tool used in physics, especially when dealing with concepts like electric potential energy. It works on scalar fields, transforming them into vector fields. Physically, this translation helps to determine the direction and rate of increase of a field's strength.

Applied to a potential energy function such as \( U(\mathbf{r}) = -q \mathbf{E}_{0} \cdot \mathbf{r} \), the gradient tells us how the energy changes with position:
  • Calculate \( abla U \) to find the force \( \mathbf{F} = - abla U \).
  • The result \( abla U = -q \mathbf{E}_{0} \) shows the force's magnitude and direction.
  • The negative sign in \( -abla U \) ensures that the force acts in the direction to reduce potential energy, characteristic of conservative forces.
Using the gradient in this way relates the abstract concept of potential energy to the more tangible notion of force. This relationship is vital for understanding how forces in fields cause motion, highlighting the predictive power of physics.

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Most popular questions from this chapter

A mass \(m\) moves in a circular orbit (centered ón the origin) in the field of an attractive central force with potential energy \(U=k r^{n} .\) Prove the virial theorem that \(T=n U / 2\).

(a) The force exerted by a one-dimensional spring, fixed at one end, is \(F=-k x,\) where \(x\) is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is \(U=\frac{1}{2} k x^{2},\) if we choose \(U\) to be zero at the equilibrium position. (b) Suppose that this spring is hung vertically from the ceiling with a mass \(m\) suspended from the other end and constrained to move in the vertical direction only. Find the extension \(x_{\mathrm{o}}\) of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form \(\frac{1}{2} k y^{2}\) if we use the coordinate \(y\) equal to the displacement measured from the new equilibrium position at \(x=x_{\mathrm{o}}\) (and redefine our reference point so that \(U=0\) at \(y=0\) ).

By writing a \(\cdot \mathbf{b}\) in terms of components prove that the product rule for differentiation applies to the dot product of two vectors; that is, \(\frac{d}{d t}(\mathbf{a} \cdot \mathbf{b})=\frac{d \mathbf{a}}{d t} \cdot \mathbf{b}+\mathbf{a} \cdot \frac{d \mathbf{b}}{d t}\).

Near to the point where I am standing on the surface of Planet \(X\), the gravitational force on a mass \(m\) is vertically down but has magnitude \(m \gamma y^{2}\) where \(\gamma\) is a constant and \(y\) is the mass's height above the horizontal ground. (a) Find the work done by gravity on a mass \(m\) moving from \(\mathbf{r}_{1}\) to \(\mathbf{r}_{2}\), and use your answer to show that gravity on Planet \(X,\) although most unusual, is still conservative. Find the corresponding potential energy. (b) Still on the same planet, I thread a bead on a curved, frictionless, rigid wire, which extends from ground level to a height \(h\) above the ground. Show clearly in a picture the forces on the bead when it is somewhere on the wire. (Just name the forces so it's clear what they are; don't worry about their magnitude.) Which of the forces are conservative and which are not? (c) If I release the bead from rest at a height \(h\), how fast will it be going when it reaches the ground?

Consider a head-on elastic collision between two particles. (since the collision is head-on, the motion is confined to a single straight line and is therefore one-dimensional.) Prove that the relative velocity after the collision is equal and opposite to that before. That is, \(v_{1}-v_{2}=-\left(v_{1}^{\prime}-v_{2}^{\prime}\right),\) where \(v_{1}\) and \(v_{2}\) are the initial velocities and \(v_{1}^{\prime}\) and \(v_{2}^{\prime}\) the corresponding final velocities.

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