Potential Energy Of An Electron
Electric potential
Voltage
Charged particles exert forces on each other. The electric field Eastward = F/q produced by a charged particle at some position r in space is a measure of the strength F the particle exerts on a exam charge q, if nosotros place the test accuse at r. The electric field Eastward is a vector. The electrical field due to a charge distribution is the vector sum of the fields produced by the charges making upward the distribution. When we recall virtually electricity in everyday life, we seldom think about the electric fields. We are more than familiar with the concepts of voltage, electric current, and power.
How is the voltage related to the electrical field?
When a particle with charge q is placed in an external electric field E, (i.e. an electric field produced by other charges), then an electric strength F = qE will act on it. If this force is non balanced past other forces, so the particle will accelerate, and its kinetic energy volition change. The electric strength volition do positive work on the particle. If the electrical force is counterbalanced by another external forcefulness F ext = -qEast, and if this external forcefulness moves the particle against the electric force, than the external forcefulness F ext does positive work.
Recall!
- The work W washed on an object by a constant force is defined equally W = F∙d = F d cosθ. Hither F is the magnitude of the force, d is the magnitude of the displacement vector, and θ is the angle between the directions of the force and deportation vectors.
The work done by a varying force in one dimension is defined as Due west = Σxi xf Fx∆x, as ∆x becomes infinitesimally small. In three dimensions we write
W = Σ F∙∆r = Σ11 xf Fx∆x + Σyi yf Fy∆y + Σzi zf Fz∆z.
Therefore the work done done by an external force balancing the electric forcefulness is
W = Σ F ext∙∆r = -q Σ E∙∆r.
In the sums we e'er presume that the displacements go infinitesimally small.
The work done by the external force F ext = -qE is equal to the change in the electrostatic potential energy of the particle in the external field. The alter in the potential energy of a charge q when existence moved from point A to point B, is the work washed past F ext in moving the charge.
∆U = UB - UA = -q ΣA B Eastward∙∆r.
The sum is taken along a particular path. Simply the electrostatic force is a conservative force. The work is independent of the path. ∆U therefore depends merely on the endpoints A and B of the path, not on the actual path itself.
Embedded Question i
(a) Which requires zero work, to motion a positive bespeak charge from indicate P to indicate one, 2, three, or 4? All those points are the same distance from P.
(b) Which requires the nigh positive work done by an external force, to move a positive point charge from point P to betoken i, ii, 3, or iv?
Talk over this with your fellow students in the discussion forum!
Review the concept of work in physics.
Discuss the relationship between the piece of work washed by and against the electric field and the electrostatic potential energy.
The change in potential free energy is proportional to the charge q. Its sign depends on whether the accuse is positive or negative. We define the potential divergence or voltage ∆V as the potential energy difference divided by the charge, or the potential free energy difference per unit (positive) charge. The potential difference is the potential energy difference of a small, positive test charge, divided by the charge. Equally in the example of gravity, the aught of the potential energy and therefore the null of the potential are not uniquely defined. Merely potential energy differences and potential differences are unique. The SI unit of energy is J, therefore the SI unit of potential is J/C. We define Volt as 5 = J/C.
The electrostatic potential is a scalar, not a vector.
We say that a accuse distribution, which produces an electric field, also produces an electric potential. The electrostatic potential produced by a finite charge distribution is, by convention, set to zero at infinity. And then the potential V(r) of the distribution is the work done per unit of measurement accuse in bringing a pocket-sized test accuse from infinity to position r. For a charge distributions which extend to infinity, we cannot set the zero of the potential at infinity, because then the potential would exist infinite everywhere, and it would exist a useless concept. Nosotros then set the zero at some convenient reference signal, but we always must specify the reference point along with the potential, since in that location is no unique convention. For many applications we cull the ground to be the cypher of the potential.
The potential at a bespeak r of a positive signal charge located at the origin is the work that must be done per unit charge in bringing a test accuse from infinity to r.
Five(r) = -Σ∞ r Eastward∙∆r = Σ r ∞ East∙∆r .
Nosotros can bring the test accuse along an arbitrary path, which we tin retrieve of as being made up of infinitesimal small steps, either in the radial management or perpendicular to the radial direction. Nosotros do not have to do work when we pace perpendicular to the radial direction, because so Due east is perpendicular to ∆r. Along the radial direction Due east∙∆r = E∆r, because E and r points both point outward. The sum becomes
5(r) = q/(4πε0r) = keq/r.
This expression likewise gives the potential due to any spherically symmetric charge distribution exterior the distribution.
The potential outside a spherically symmetric charge distribution with total accuse q is the same as that of a bespeak charge q, V(r) = grandeq/r.
Information technology is is proportional to the changed of the distance from the indicate charge.
To find the potential due to a collection of charges, nosotros apply the principle of superposition and add together the potentials due to the private charges. Because the potential is a scalar, and not a vector, we just have to add numbers. To find the total electrostatic potential free energy of a collection of point charges, sum over all pairs.
Trouble:
Solution:
- Reasoning:
For example A:
We have only 1 pair of charges.
W = keastward 2Q*Q/d = two 1000eQii/d.For case B:
We take 3 pairs of charges.
Bring the two lower charges together: W = kdue east Q2/d.
Bring the elevation charge:
Do work against the forcefulness exerted past the two lower charges
W = ke Q2/d + ge Qtwo/d = two ke Qii/d.
Full Work: West = 3 keastwardQ2/d = sum over all pairs.Grouping A took less work by an external force to bring together from infinity than grouping B.
Units:
1 electron volt (eV) = 1.vi*10-xix J.
1 eV is the modify in potential energy of a particle with charge qdue east = 1.half dozen*ten-9 C when the change in potential is 1 Volt (V). The unit was divers and so that when yous know the voltage between 2 points in space, you know the change in potential energy of an uncomplicated particle when it moves from one to the other point. (The sign of the change in potential free energy depends on the sign of the charge.) When a complimentary proton moves through a potential deviation of 1 Five its kinetic energy decreases by -qV = (1.6*10-19 C)*(i J/C) = -1.6*10-nineteen J = -i eV. When a free electron moves through the aforementioned potential divergence of 1 V its kinetic energy increases past -qV = -(-i.6*x-19 C)*(one J/C) = 1.6*10-19 J = 1 eV.
Problem:
Solution:
- Reasoning:
The electrostatic strength is a conservative force.
∆KE + ∆U = 0.
(modify in kinetic energy + alter in potential energy = 0.) - Details of the adding:
The proton accelerates towards the negative plate. Its potential free energy decreases by 100 eV.
Its kinetic energy increases by 1.6*10-17 J = 100 eV.
Summary:
By definition, the potential divergence is the potential energy departure of a small, positive test charge, divided past the charge. The zero of the potential free energy and therefore of the potential is not uniquely defined, but chosen at a user-friendly place. Voltage is just some other discussion for potential difference, or potential free energy difference divided by the charge.
∆V = ∆U/q.
A particle ever accelerates towards the position with the lower potential energy U. The electrostatic potential free energy is U = qV, were Five is the potential. Consider a particle with accuse of magnitude qeastward, for example a proton (+qe) or and electron (-qdue east).
If q = qe, and so U = qeV. U gets more than positive or higher, the bigger V. The positively charged particle accelerates towards the region of lower potential.
If q = -qeast, so U = -qeFive. U gets more negative or lower, the bigger 5. The negatively charged particle accelerates towards the region of higher potential.
If the potential difference is 5, the change in potential free energy of the particle U can be calculated in SI units. Only the nice thing about the unit eV is, that if the voltage departure is given in volts V, and so the alter in the potential energy of the particle, ±qeU, is just the same number in units of eV.
Equipotential lines
Just equally we described the electric field around a charged object by field lines, we tin as well describe the electric potential pictorially with equipotential surfaces (contour plots). Each surface corresponds to a different fixed value of the potential. Equipotential lines are lines connecting points of the same potential. They frequently appear on field line diagrams.
Equipotential lines are always perpendicular to field lines.
The electric potential at a position r is the electric potential energy a accuse q has if it is at position r. As the name equipotential implies, the potential energy of a charge would be the aforementioned anywhere on an equipotential surface. If the charge moved along an equipotential surface, its electric potential energy would not change, no work would exist done past or confronting the electric field. Only if there was a component of the electric field tangent to the equipotential surface, then, if the charge moved along that component, the electric field would exercise work and the potential free energy of the accuse would change. This is a contradiction. We cannot have a component of the electric field tangent to an equipotential surface. Hence, whatsoever electric field must be perpendicular to any equipotential surface.
Instance:
Field lines and equipotential lines for a positive betoken accuse are shown in the figure. This is a two-dimensional representation, a cut through the 3-d surfaces.
Delight also explore this 3-dimensional representation below. Delight click on the paradigm!
Problem:
In the Bohr model of the hydrogen atom, the electron orbits the proton at a altitude of r = 5.29*ten-11 m. The proton has charge +qeast and the electron has charge -qeastward, where qe = one.6*10-19 C. How much piece of work must exist done to completely separate the electron and the proton
Solution:
- Reasoning:
The force on the electron is the Coulomb forcefulness between the proton and the electron. It pulls the electron towards the proton. For the electron to move in a circular orbit, the Coulomb force must equal the centripetal force. We need yardeastwardqe ii/r2 = mv2/r. - Details of the calculation:
mv2 = keqe 2/r, and then the kinetic energy of the electron is
KE(r) = ½mvii = ½thoueqeastward 2/r.
The potential energy of the electron in the field of the positive proton point accuse is U(r) = -qeV(r) = - keqe ii/r.
The total energy is the sum of the electron'due south kinetic energy and its potential energy.
KE(r) + PE(r) = -½thoueqeast ii/r = (-½) (9*x9)(one.60*10-19) /(v.29*10-11) J = -2.eighteen*10-18 J.
This is normally stated in free energy units of electron volts (eV). 1 eV = 1.60*10-19 J.
-two.18*10-18 J * 1eV/(1.threescore*10-19 J) = -13.6 eV.
To remove the electron from the cantlet, i.e. to motility information technology very far away and give it nil kinetic energy, 13.6 eV of piece of work must exist done past an external strength.
13.half-dozen eV is the ionization energy of hydrogen.
Trouble:
An alpha particle containing two protons is shot straight towards a platinum nucleus containing 78 protons from a very big altitude with a kinetic free energy of 1.vii*10-12 J. What will be the altitude of closest approach?
Solution:
- Reasoning:
The electric charge of the blastoff particle is qone = 2qe and that of the platinum nucleus is q2 = 78 qdue east. The alpha particle and the nucleus repel each other. As the alpha particle moves towards the nucleus, some of its kinetic energy will be converted into electrostatic potential energy. At the altitude of closest approach, the blastoff particle's velocity is zero, and all its initial kinetic free energy has been converted into electrostatic potential energy. - Details of the calculation:
The initial kinetic energy of the alpha particle is KE = 1.7*10-12 J.
The final potential energy is U = kqaneq2/d, where d is the altitude of closest approach.
We have keq1q2/d = one.7*10-12 J. This yields
d = chiliadeqiq2/(1.7*ten-12 J) = (nine*10ix*2*78*(i.6*10-19)2/(1.vii*ten-12)) k = ii.1*ten-14 m
for the distance of closest approach. The blastoff particle's velocity is zero at d.
The acceleration is in a management away from the nucleus. The distance between the alpha particle and the nucleus volition increment again, converting the potential energy back into kinetic free energy.
Trouble:
An evacuated tube uses a voltage of 5 kV to accelerate electrons from remainder to striking a phosphor screen.
(a) How much kinetic energy does each electron gain?
(b) What is the speed of an electron when it hits the copper plate?
Solution:
- Reasoning:
1 eV is the change in potential energy of a particle with charge qe = 1.half-dozen*x-9 C when the change in potential is 1 Volt (V). - Details of the calculation:
(a) Each electron loses v keV of potential energy and gains 5 keV = (5000 eV)(1.6*10-xix J/eV) = eight*10-16 J of kinetic energy.
(b) Eastward = ½mefive2. vtwo = 2E/me = (2*8*10-16 J)/(9.1*ten-31 kg) = 1.75*x15 (yard/s)two.
5 = iv.2*ten7 m/due south. This is more than 1/10 the speed of light.
The electrostatic potential V is related to the electrostatic field East. If the electric field E is known, the electrostatic potential V tin can be obtained using V(r) = -Σ∞ r Eastward∙∆r.
How can we obtain the field from the potential?
Consider the two points Pane and P2 shown in figure higher up. Presume that they are separated by an infinitesimal distance ∆L. The change in the electrostatic potential betwixt P1 and Ptwo is given by ∆V = -Due east∙∆L = -East ∆Fifty cosθ. Here θ is the angle between the direction of the electrical field and the direction of the deportation vector ∆L. We can rewrite this equation every bit
∆Five/∆L = -E cosθ = -Due east50.
EL indicates the component of the electric field along the management of ∆L. If the direction of the displacement is called to coincide with the x-axis, this becomes
∆Five/∆10 = -E10.
For the displacements along the y-centrality and z-axis we obtain ∆V/∆y = -Ey and ∆V/∆z = -Ez.
The total electrical field E can be obtained from the electrostatic potential Five by combining these three equations.
We say that E is the negative gradient of the potential V.
In many electrostatic problems the electrical field due to a certain charge distribution must be evaluated. The calculation of the electrical field tin be carried out using two different methods:
- The electric field can be calculated by applying Coulomb's constabulary and vector addition of the contributions from all charges in the accuse distribution.
- The total electrostatic potential 5 can be obtained from the algebraic sum of the potential due to all charges that make up the charge distribution, and the field can be found by calculating the slope of 5.
In many cases the second method is simpler, because the calculation of the electrostatic potential involves an algebraic sum, while the direct adding of the field involves a vector sum.
Trouble:
The potential deviation between the two plates of the capacitor shown below is 12 5. Equipotential surfaces are shown. If the separation between the plates is 1 mm, what is the forcefulness of the electric field between the plates?
Solution:
- Reasoning:
Let the y-axis indicate upwardly. 5 only varies with y. ∆V/∆x = ∆5/∆z = 0. Eastward = Ey j = -∆V/∆y j. - Details of the calculation:
Let the y-axis point upward. The field is uniform and points in the negative y-management.
∆V/∆y = -Ey = (12 V)/(x-3 m). Eastwardy = -12000 Five/one thousand.
The strength of the electrical field between the plates is E = 12000 5/one thousand = 12000 N/C.
Problem:
Solution:
- Reasoning:
Point P lies betwixt ii equipotential surfaces. The surface to a higher place P is at 40 V and the surface below is at 20 Five. The electric field points abroad from the positive charge, from college to lower potential. It points downwards at P.
Given the distance scale, I guess the perpendicular distance between the 40 V and 20 Five equipontential surfaces near signal P to be ii.5 cm. - Details of the calculation:
|E| = ∆V/∆y = 20 5/ 2.5*10-two m = 800 V/m.
The magnitude on of the electric field at point P is ~800 V/m and it points downwardly.
If you miss having regular lectures, consider this video lecture
Lecture iv: Electrostatic Potential, Electric Energy, eV, Conservative Field, Equipotential Surfaces
Potential Energy Of An Electron,
Source: http://labman.phys.utk.edu/phys222core/modules/m2/Electric%20potential.html#:~:text=The%20potential%20energy%20of%20the,energy%20and%20its%20potential%20energy.
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