What Is The Change In The Potential Energy Of Charge +q During This Process?
Affiliate vii. Electric Potential
seven.1 Electric Potential Energy
Learning Objectives
Past the finish of this section, you will be able to:
- Define the piece of work washed by an electric force
- Define electric potential free energy
- Utilize work and potential free energy in systems with electric charges
When a complimentary positive charge q is accelerated past an electric field, it is given kinetic energy (Figure 7.2). The process is analogous to an object being accelerated by a gravitational field, as if the accuse were going downwardly an electrical hill where its electric potential free energy is converted into kinetic free energy, although of grade the sources of the forces are very different. Let us explore the work washed on a charge q by the electric field in this procedure, so that we may develop a definition of electric potential energy.
The electrostatic or Coulomb force is conservative, which ways that the work done on q is independent of the path taken, as we will demonstrate afterwards. This is exactly analogous to the gravitational strength. When a force is conservative, it is possible to ascertain a potential free energy associated with the force. It is normally easier to work with the potential free energy (because it depends only on position) than to calculate the work directly.
To bear witness this explicitly, consider an electrical accuse [latex]+q[/latex] fixed at the origin and movement another accuse [latex]+Q[/latex] toward q in such a manner that, at each instant, the applied force [latex]\stackrel{\to }{\textbf{F}}[/latex] exactly balances the electric strength [latex]{\stackrel{\to }{\textbf{F}}}_{e}[/latex] on Q (Effigy 7.3). The work done by the applied force [latex]\stackrel{\to }{\textbf{F}}[/latex] on the charge Q changes the potential energy of Q. Nosotros call this potential energy the electrical potential free energy of Q.
The work [latex]{W}_{12}[/latex] done past the applied strength [latex]\stackrel{\to }{\textbf{F}}[/latex] when the particle moves from [latex]{P}_{1}[/latex] to [latex]{P}_{2}[/latex] may be calculated by
[latex]{W}_{12}={\int }_{{P}_{1}}^{{P}_{2}}\stackrel{\to }{\textbf{F}}·d\stackrel{\to }{\textbf{l}}.[/latex]
Since the applied force [latex]\stackrel{\to }{\textbf{F}}[/latex] balances the electric force [latex]{\stackrel{\to }{\textbf{F}}}_{e}[/latex] on Q, the two forces have equal magnitude and contrary directions. Therefore, the practical force is
[latex]\stackrel{\to }{\textbf{F}}=\text{−}\stackrel{\to }{\textbf{F}_{due east}}=-\frac{kqQ}{{r}^{2}}\hat{\textbf{r}},[/latex]
where we take defined positive to exist pointing abroad from the origin and r is the distance from the origin. The directions of both the displacement and the applied force in the system in Effigy 7.iii are parallel, and thus the piece of work washed on the system is positive.
Nosotros use the letter U to announce electric potential energy, which has units of joules (J). When a conservative force does negative work, the system gains potential energy. When a conservative forcefulness does positive work, the system loses potential energy, [latex]\text{Δ}U=\text{−}W.[/latex] In the system in Figure 7.3, the Coulomb force acts in the reverse direction to the displacement; therefore, the work is negative. Notwithstanding, we have increased the potential free energy in the two-charge system.
Example
Kinetic Energy of a Charged Particle
A [latex]+3.0\text{-nC}[/latex] charge Q is initially at remainder a distance of 10 cm ([latex]{r}_{1}[/latex]) from a [latex]+v.0\text{-nC}[/latex] accuse q fixed at the origin (Figure 7.4). Naturally, the Coulomb force accelerates Q away from q, eventually reaching 15 cm ([latex]{r}_{two}[/latex]).
- What is the work washed by the electric field betwixt [latex]{r}_{i}[/latex] and [latex]{r}_{ii}[/latex]?
- How much kinetic free energy does Q have at [latex]{r}_{ii}[/latex]?
Strategy
Calculate the work with the usual definition. Since Q started from rest, this is the same as the kinetic free energy.
Solution
Prove Answer
Integrating force over altitude, we obtain
[latex]\begin{array}{cc}{West}_{12}\hfill & ={\int }_{{r}_{1}}^{{r}_{ii}}\stackrel{\to }{\textbf{F}}·d\stackrel{\to }{\textbf{r}}={\int }_{{r}_{one}}^{{r}_{2}}\frac{kqQ}{{r}^{two}}\phantom{\rule{0.2em}{0ex}}dr={\left[-\frac{kqQ}{r}\correct]}_{{r}_{one}}^{{r}_{two}}=kqQ\left[\frac{-one}{{r}_{ii}}+\frac{one}{{r}_{1}}\right]\hfill \\ & =\left(eight.99\phantom{\dominion{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}{\text{Nm}}^{2}{\text{/C}}^{2}\right)\left(5.0\phantom{\dominion{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-nine}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{ten}^{-nine}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left[\frac{-1}{0.15\phantom{\dominion{0.2em}{0ex}}\text{one thousand}}+\frac{ane}{0.x\phantom{\rule{0.2em}{0ex}}\text{yard}}\correct]\hfill \\ & =iv.5\phantom{\rule{0.2em}{0ex}}×\phantom{\dominion{0.2em}{0ex}}{10}^{-seven}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}\hfill \cease{array}[/latex]
This is also the value of the kinetic energy at [latex]{r}_{2}.[/latex]
Significance
Accuse Q was initially at rest; the electric field of q did work on Q, so at present Q has kinetic energy equal to the work done by the electrical field.
Check Your Understanding
If Q has a mass of [latex]iv.00\phantom{\rule{0.2em}{0ex}}\mu \text{g},[/latex] what is the speed of Q at [latex]{r}_{two}?[/latex]
Evidence Solution
[latex]1000=\frac{1}{2}\phantom{\rule{0.2em}{0ex}}g{v}^{2},\phantom{\rule{0.2em}{0ex}}v=\sqrt{2\frac{K}{m}}=\sqrt{two\frac{4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{x}^{-7}\phantom{\rule{0.2em}{0ex}}\text{J}}{four.00\phantom{\rule{0.2em}{0ex}}×\phantom{\dominion{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}\text{kg}}}=15\phantom{\rule{0.2em}{0ex}}\text{m/s}[/latex]
In this case, the piece of work W washed to accelerate a positive accuse from rest is positive and results from a loss in U, or a negative [latex]\text{Δ}U[/latex]. A value for U can be found at whatever betoken by taking i point every bit a reference and computing the work needed to move a charge to the other point.
Electric Potential Energy
Work Westward done to accelerate a positive accuse from rest is positive and results from a loss in U, or a negative [latex]\text{Δ}U[/latex]. Mathematically,
[latex]Westward=\text{−}\text{Δ}U.[/latex]
Gravitational potential energy and electric potential free energy are quite analogous. Potential energy accounts for work done by a bourgeois force and gives added insight regarding energy and energy transformation without the necessity of dealing with the force directly. Information technology is much more than common, for instance, to use the concept of electric potential energy than to deal with the Coulomb force directly in real-world applications.
In polar coordinates with q at the origin and Q located at r, the displacement element vector is [latex]d\stackrel{\to }{\textbf{l}}=\hat{\textbf{r}}\phantom{\rule{0.2em}{0ex}}dr[/latex] and thus the work becomes
[latex]{W}_{12}=\text{−}kqQ{\int }_{{r}_{i}}^{{r}_{2}}\frac{1}{{r}^{ii}}\hat{\textbf{r}}·\chapeau{\textbf{r}}dr=kqQ\frac{i}{{r}_{two}}-kqQ\frac{one}{{r}_{1}}.[/latex]
Notice that this issue only depends on the endpoints and is otherwise contained of the path taken. To explore this farther, compare path [latex]{P}_{1}[/latex] to [latex]{P}_{ii}[/latex] with path [latex]{P}_{1}{P}_{3}{P}_{4}{P}_{2}[/latex] in Figure vii.5.
The segments [latex]{P}_{i}{P}_{3}[/latex] and [latex]{P}_{iv}{P}_{two}[/latex] are arcs of circles centered at q. Since the force on Q points either toward or away from q, no work is done by a strength balancing the electric strength, because it is perpendicular to the deportation along these arcs. Therefore, the only piece of work done is along segment [latex]{P}_{3}{P}_{4},[/latex] which is identical to [latex]{P}_{1}{P}_{2}.[/latex]
One implication of this work calculation is that if we were to go around the path [latex]{P}_{1}{P}_{3}{P}_{4}{P}_{2}{P}_{ane},[/latex] the cyberspace work would be nothing (Figure 7.six). Recall that this is how nosotros determine whether a force is conservative or non. Hence, considering the electric forcefulness is related to the electric field by [latex]\stackrel{\to }{\textbf{F}}=q\stackrel{\to }{\textbf{E}}[/latex], the electric field is itself conservative. That is,
[latex]\oint \stackrel{\to }{\textbf{E}}·d\stackrel{\to }{\textbf{l}}=0.[/latex]
Notation that Q is a abiding.
Another implication is that we may define an electric potential energy. Recall that the work done by a conservative forcefulness is also expressed as the difference in the potential energy corresponding to that force. Therefore, the work [latex]{W}_{\text{ref}}[/latex] to bring a accuse from a reference point to a indicate of interest may be written as
[latex]{West}_{\text{ref}}={\int }_{{r}_{\text{ref}}}^{r}\stackrel{\to }{\textbf{F}}·d\stackrel{\to }{\textbf{50}}[/latex]
and, by Equation 7.1, the difference in potential free energy [latex]\left({U}_{2}-{U}_{1}\correct)[/latex] of the exam accuse Q betwixt the 2 points is
[latex]\text{Δ}U=\text{−}{\int }_{{r}_{\text{ref}}}^{r}\stackrel{\to }{\textbf{F}}·d\stackrel{\to }{\textbf{l}}.[/latex]
Therefore, nosotros tin can write a general expression for the potential free energy of 2 signal charges (in spherical coordinates):
[latex]\text{Δ}U=\text{−}{\int }_{{r}_{\text{ref}}}^{r}\frac{kqQ}{{r}^{two}}dr=\text{−}{\left[-\frac{kqQ}{r}\right]}_{{r}_{\text{ref}}}^{r}=kqQ\left[\frac{1}{r}-\frac{1}{{r}_{\text{ref}}}\right].[/latex]
We may have the 2d term to be an arbitrary abiding reference level, which serves as the nix reference:
[latex]U\left(r\right)=k\frac{qQ}{r}-{U}_{\text{ref}}.[/latex]
A convenient selection of reference that relies on our common sense is that when the 2 charges are infinitely far apart, in that location is no interaction between them. (Recall the give-and-take of reference potential energy in Potential Free energy and Conservation of Energy.) Taking the potential energy of this country to be nix removes the term [latex]{U}_{\text{ref}}[/latex] from the equation (just like when we say the basis is nix potential energy in a gravitational potential energy problem), and the potential free energy of Q when information technology is separated from q by a distance r assumes the grade
[latex]U\left(r\right)=thousand\frac{qQ}{r}\phantom{\rule{0.2em}{0ex}}\left(z\text{ero reference at}\phantom{\dominion{0.2em}{0ex}}r=\infty \right).[/latex]
This formula is symmetrical with respect to q and Q, and so it is all-time described equally the potential free energy of the 2-charge organization.
Example
Potential Energy of a Charged Particle
A [latex]+3.0\text{-nC}[/latex] charge Q is initially at rest a altitude of x cm ([latex]{r}_{one}[/latex]) from a [latex]\text{+v}\text{.0-nC}[/latex] charge q fixed at the origin (Figure 7.seven). Naturally, the Coulomb force accelerates Q away from q, somewhen reaching fifteen cm ([latex]{r}_{2}[/latex]).
What is the modify in the potential energy of the two-accuse system from [latex]{r}_{1}[/latex] to [latex]{r}_{ii}?[/latex]
Strategy
Calculate the potential energy with the definition given in a higher place: [latex]\text{Δ}{U}_{12}=\text{−}{\int }_{{r}_{1}}^{{r}_{ii}}\stackrel{\to }{\textbf{F}}·d\stackrel{\to }{\textbf{r}}.[/latex] Since Q started from residue, this is the same every bit the kinetic free energy.
Solution
Show Answer
We have
[latex]\brainstorm{array}{cc}\text{Δ}{U}_{12}\hfill & =\text{−}{\int }_{{r}_{1}}^{{r}_{2}}\stackrel{\to }{\textbf{F}}·d\stackrel{\to }{\textbf{r}}=\text{−}{\int }_{{r}_{ane}}^{{r}_{2}}\frac{kqQ}{{r}^{2}}dr=\text{−}{\left[-\frac{kqQ}{r}\right]}_{{r}_{1}}^{{r}_{2}}=kqQ\left[\frac{1}{{r}_{2}}-\frac{ane}{{r}_{1}}\right]\hfill \\ & =\left(8.99\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}{\text{Nm}}^{ii}{\text{/C}}^{two}\right)\left(5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-nine}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{ten}^{-9}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left[\frac{1}{0.15\phantom{\rule{0.2em}{0ex}}\text{m}}-\frac{1}{0.10\phantom{\rule{0.2em}{0ex}}\text{m}}\correct]\hfill \\ & =-4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-seven}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}\hfill \end{array}[/latex]
Significance
The modify in the potential energy is negative, every bit expected, and equal in magnitude to the change in kinetic energy in this organisation. Recall from Instance 7.ane that the change in kinetic energy was positive.
Check Your Agreement
What is the potential energy of Q relative to the zilch reference at infinity at [latex]{r}_{2}[/latex] in the above example?
Evidence Solution
It has kinetic energy of [latex]iv.v\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{x}^{-7}\phantom{\dominion{0.2em}{0ex}}\text{J}[/latex] at signal [latex]{r}_{ii}[/latex] and potential energy of [latex]9.0\phantom{\dominion{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-vii}\phantom{\rule{0.2em}{0ex}}\text{J},[/latex] which means that as Q approaches infinity, its kinetic energy totals three times the kinetic energy at [latex]{r}_{2},[/latex] since all of the potential free energy gets converted to kinetic.
Due to Coulomb'south constabulary, the forces due to multiple charges on a test charge Q superimpose; they may be calculated individually and so added. This implies that the piece of work integrals and hence the resulting potential energies exhibit the same behavior. To demonstrate this, nosotros consider an case of assembling a system of four charges.
Example
Assembling Four Positive Charges
Notice the corporeality of piece of work an external agent must exercise in assembling 4 charges [latex]+2.0\phantom{\rule{0.2em}{0ex}}\mu \text{C},[/latex][latex]+3.0\phantom{\rule{0.2em}{0ex}}\mu \text{C},\phantom{\rule{0.2em}{0ex}}+4.0\phantom{\dominion{0.2em}{0ex}}\mu \text{C},[/latex] and [latex]+5.0\phantom{\dominion{0.2em}{0ex}}\mu \text{C}[/latex] at the vertices of a square of side ane.0 cm, starting each accuse from infinity (Figure 7.8).
Strategy
Nosotros bring in the charges 1 at a time, giving them starting locations at infinity and computing the work to bring them in from infinity to their final location. We exercise this in society of increasing accuse.
Solution
Show Reply
Step 1. First bring the [latex]+2.0\text{-}\mu \text{C}[/latex] charge to the origin. Since at that place are no other charges at a finite distance from this charge nonetheless, no work is done in bringing information technology from infinity,
[latex]{Westward}_{one}=0.[/latex]
Step ii. While keeping the [latex]+ii.0\text{-}\mu \text{C}[/latex] charge stock-still at the origin, bring the [latex]+3.0\text{-}\mu \text{C}[/latex] charge to [latex]\left(x,y,z\correct)=\left(1.0\phantom{\dominion{0.2em}{0ex}}\text{cm},0,0\right)[/latex] (Figure seven.nine). At present, the practical forcefulness must do work against the force exerted past the [latex]+2.0\text{-}\mu \text{C}[/latex] charge fixed at the origin. The work done equals the alter in the potential energy of the [latex]+3.0\text{-}\mu \text{C}[/latex] charge:
[latex]{W}_{two}=grand\frac{{q}_{1}{q}_{2}}{{r}_{12}}=\left(9.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{ten}^{nine}\frac{\text{N}·{\text{chiliad}}^{2}}{{\text{C}}^{2}}\right)\frac{\left(ii.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{ten}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\correct)\left(3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-half-dozen}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-ii}\phantom{\rule{0.2em}{0ex}}\text{m}}=five.iv\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}[/latex]
Step iii. While keeping the charges of [latex]+ii.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}[/latex] and [latex]+iii.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}[/latex] fixed in their places, bring in the [latex]+4.0\text{-}\mu \text{C}[/latex] accuse to [latex]\left(ten,y,z\right)=\left(1.0\phantom{\dominion{0.2em}{0ex}}\text{cm},1.0\phantom{\rule{0.2em}{0ex}}\text{cm},0\correct)[/latex] (Effigy seven.10). The piece of work washed in this step is
[latex]\brainstorm{array}{cc}{W}_{3}\hfill & =thousand\frac{{q}_{1}{q}_{three}}{{r}_{13}}+chiliad\frac{{q}_{two}{q}_{3}}{{r}_{23}}\hfill \\ & =\left(9.0\phantom{\dominion{0.2em}{0ex}}×\phantom{\dominion{0.2em}{0ex}}{10}^{nine}\frac{\text{Northward}·{\text{m}}^{2}}{{\text{C}}^{ii}}\right)\left[\frac{\left(2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\dominion{0.2em}{0ex}}\text{C}\right)\left(4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\dominion{0.2em}{0ex}}{x}^{-vi}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{\sqrt{2}\phantom{\dominion{0.2em}{0ex}}×\phantom{\dominion{0.2em}{0ex}}{10}^{-two}\phantom{\rule{0.2em}{0ex}}\text{m}}+\frac{\left(3.0\phantom{\dominion{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{x}^{-half dozen}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(iv.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{ten}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{1.0\phantom{\dominion{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{x}^{-2}\phantom{\rule{0.2em}{0ex}}\text{thousand}}\right]=fifteen.9\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}\hfill \end{array}[/latex]
Pace four. Finally, while keeping the kickoff three charges in their places, bring the [latex]+v.0\text{-}\mu \text{C}[/latex] charge to [latex]\left(ten,y,z\right)=\left(0,\phantom{\rule{0.2em}{0ex}}1.0\phantom{\rule{0.2em}{0ex}}\text{cm},\phantom{\rule{0.2em}{0ex}}0\correct)[/latex] (Figure seven.11). The piece of work done here is
[latex]\begin{array}{cc}{West}_{4}\hfill & =yard{q}_{4}\left[\frac{{q}_{1}}{{r}_{fourteen}}+\frac{{q}_{2}}{{r}_{24}}+\frac{{q}_{3}}{{r}_{34}}\right],\hfill \\ & =\left(ix.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\frac{\text{N}·{\text{m}}^{2}}{{\text{C}}^{2}}\right)\left(five.0\phantom{\dominion{0.2em}{0ex}}×\phantom{\dominion{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left[\frac{\left(2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\dominion{0.2em}{0ex}}{x}^{-six}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-ii}\phantom{\dominion{0.2em}{0ex}}\text{m}}+\frac{\left(3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\dominion{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{\sqrt{ii}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\dominion{0.2em}{0ex}}\text{m}}+\frac{\left(4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{x}^{-vi}\phantom{\dominion{0.2em}{0ex}}\text{C}\right)}{one.0\phantom{\rule{0.2em}{0ex}}×\phantom{\dominion{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{m}}\correct]=36.5\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}\hfill \end{assortment}[/latex]
Hence, the full work done by the applied force in assembling the 4 charges is equal to the sum of the piece of work in bringing each accuse from infinity to its final position:
[latex]{W}_{\text{T}}={W}_{1}+{W}_{2}+{W}_{3}+{W}_{4}=0+5.4\phantom{\rule{0.2em}{0ex}}\text{J}+15.nine\phantom{\dominion{0.2em}{0ex}}\text{J}+36.v\phantom{\rule{0.2em}{0ex}}\text{J}=57.eight\phantom{\dominion{0.2em}{0ex}}\text{J}\text{.}[/latex]
Significance
The work on each charge depends only on its pairwise interactions with the other charges. No more complicated interactions need to be considered; the work on the tertiary accuse only depends on its interaction with the first and 2nd charges, the interaction between the outset and second charge does not affect the tertiary.
Check Your Understanding
Is the electrical potential energy of ii bespeak charges positive or negative if the charges are of the same sign? Opposite signs? How does this relate to the work necessary to bring the charges into proximity from infinity?
Bear witness Solution
positive, negative, and these quantities are the same every bit the work y'all would demand to practice to bring the charges in from infinity
Note that the electrical potential energy is positive if the two charges are of the same type, either positive or negative, and negative if the two charges are of opposite types. This makes sense if yous think of the modify in the potential free energy [latex]\text{Δ}U[/latex] every bit you bring the two charges closer or move them farther apart. Depending on the relative types of charges, you may take to piece of work on the system or the system would do work on you, that is, your work is either positive or negative. If y'all accept to practise positive work on the system (actually push the charges closer), then the energy of the arrangement should increase. If you bring 2 positive charges or two negative charges closer, you lot have to do positive piece of work on the system, which raises their potential energy. Since potential energy is proportional to ane/r, the potential energy goes up when r goes down between two positive or two negative charges.
On the other hand, if you bring a positive and a negative charge nearer, you accept to exercise negative work on the system (the charges are pulling y'all), which means that yous accept energy away from the system. This reduces the potential energy. Since potential free energy is negative in the case of a positive and a negative accuse pair, the increase in 1/r makes the potential free energy more negative, which is the same as a reduction in potential energy.
The issue from Example vii.1 may be extended to systems with any arbitrary number of charges. In this instance, it is nigh convenient to write the formula equally
[latex]{W}_{12\cdots N}=\frac{m}{two}\sum _{i}^{Due north}\sum _{j}^{North}\frac{{q}_{i}{q}_{j}}{{r}_{ij}}\phantom{\dominion{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}i\ne j.[/latex]
The factor of 1/2 accounts for adding each pair of charges twice.
Summary
- The piece of work washed to move a accuse from betoken A to B in an electric field is path independent, and the work around a closed path is zero. Therefore, the electric field and electric force are bourgeois.
- We can ascertain an electric potential free energy, which between bespeak charges is [latex]U\left(r\right)=1000\frac{qQ}{r}[/latex], with the zilch reference taken to exist at infinity.
- The superposition principle holds for electric potential energy; the potential energy of a organisation of multiple charges is the sum of the potential energies of the individual pairs.
Conceptual Questions
Would electric potential energy be meaningful if the electric field were not conservative?
Show Solution
No. We can merely define potential energies for conservative fields.
Why do nosotros demand to be careful about work done on the organisation versus work done by the system in calculations?
Does the gild in which we assemble a system of indicate charges touch the total piece of work done?
Bear witness Solution
No, though certain orderings may be simpler to compute.
Problems
Consider a charge [latex]{Q}_{1}\left(+v.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}\right)[/latex] fixed at a site with another charge [latex]{Q}_{2}[/latex] (accuse [latex]+3.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}[/latex], mass [latex]6.0\phantom{\rule{0.2em}{0ex}}\mu \text{g}[/latex]) moving in the neighboring space. (a) Evaluate the potential energy of [latex]{Q}_{2}[/latex] when it is 4.0 cm from [latex]{Q}_{1}.[/latex] (b) If [latex]{Q}_{2}[/latex] starts from remainder from a point four.0 cm from [latex]{Q}_{1},[/latex] what will be its speed when it is eight.0 cm from [latex]{Q}_{one}[/latex]? (Note: [latex]{Q}_{one}[/latex] is held stock-still in its place.)
Evidence Solution
a. [latex]U=3.4\phantom{\rule{0.2em}{0ex}}\text{J;}[/latex]
b. [latex]\frac{i}{2}chiliad{5}^{ii}={Q}_{1}{Q}_{2}\left(\frac{1}{{r}_{i}}-\frac{one}{{r}_{f}}\right)\to v=2.4x{ten}^{4}\phantom{\rule{0.2em}{0ex}}\text{m/s}[/latex]
Two charges [latex]{Q}_{one}\left(+2.00\phantom{\rule{0.2em}{0ex}}\mu \text{C}\correct)[/latex] and [latex]{Q}_{two}\left(+two.00\phantom{\rule{0.2em}{0ex}}\mu \text{C}\right)[/latex] are placed symmetrically along the x-axis at [latex]x=±3.00\phantom{\rule{0.2em}{0ex}}\text{cm}[/latex]. Consider a charge [latex]{Q}_{3}[/latex] of charge [latex]+four.00\phantom{\rule{0.2em}{0ex}}\mu \text{C}[/latex] and mass 10.0 mg moving along the y-axis. If [latex]{Q}_{3}[/latex] starts from remainder at [latex]y=2.00\phantom{\dominion{0.2em}{0ex}}\text{cm,}[/latex] what is its speed when information technology reaches [latex]y=4.00\phantom{\rule{0.2em}{0ex}}\text{cm?}[/latex]
To form a hydrogen atom, a proton is stock-still at a point and an electron is brought from far abroad to a altitude of [latex]0.529\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\phantom{\dominion{0.2em}{0ex}}\text{m,}[/latex] the average distance between proton and electron in a hydrogen atom. How much work is done?
Evidence Solution
[latex]U=four.36\phantom{\rule{0.2em}{0ex}}×\phantom{\dominion{0.2em}{0ex}}{10}^{-eighteen}\phantom{\dominion{0.2em}{0ex}}\text{J}\phantom{\rule{0.2em}{0ex}}[/latex]
(a) What is the average power output of a center defibrillator that dissipates 400 J of free energy in x.0 ms? (b) Considering the high-power output, why doesn't the defibrillator produce serious burns?
Glossary
- electric potential energy
- potential free energy stored in a arrangement of charged objects due to the charges
Licenses and Attributions
Electric Potential Energy. Authored by: OpenStax College. Located at: https://openstax.org/books/university-physics-volume-ii/pages/vii-1-electric-potential-free energy. License: CC BY: Attribution. License Terms: Download for free at https://openstax.org/books/university-physics-book-2/pages/1-introduction
Source: https://pressbooks.online.ucf.edu/osuniversityphysics2/chapter/electric-potential-energy/
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