As friend recall, we defined the electric field, $$\vec E(\vec r)$$, to be the electric pressure per unit charge. By defining an electrical field almost everywhere in space, we were maybe to quickly determine the pressure on any test charge, $$q$$, whether the test fee is hopeful or an unfavorable (since the sign of $$q$$ will readjust the direction that the pressure vector, $$q\vec E$$): \<\beginaligned \vec E(\vec r) &= \frac\vec F^E(\vec r)q\\ \therefore \vec F^E(\vec r)&=q\vec E(\vec r)\endaligned\> Similarly, we define the electric potential, $$V(\vec r)$$, to it is in the electric potential energy per unit charge. This enables us to define electric potential, $$V(\vec r)$$, all over in space, and also then identify the potential energy of a particular charge, $$q$$, by just multiplying $$q$$ with the electrical potential at that position in space. \<\beginaligned V(\vec r) &= \frac U(\vec r)q\\ \therefore U(\vec r)&= q V(\vec r)\endaligned\> The S.I. Unit for electrical potential is the “volt”, (V). Electrical potential, $$V(\vec r)$$, is a scalar field whose value is “the electrical potential” at that position in space. A hopeful charge, $$q=1\textC$$, will certainly thus have actually a potential energy of $$U=10\textJ$$ if the is situated at a position in room where the electric potential is $$V=10\textV$$, since $$U=qV$$. Similarly, a negative charge, $$q=-1\textC$$, will have an adverse potential energy, $$U=-10\textJ$$, in ~ the same location.

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Since only differences in potential power are londonchinatown.orgically coherent (as change in potential energy is concerned work), only transforms in electric potential are londonchinatown.orgically meaningful (as electric potential is associated to electrical potential energy). A difference in electric potential is generally called a “voltage”. One regularly makes a clear an option of whereby the electric potential is zero (typically the ground, or infinitely much away), so the the ax voltage is used to explain potential, $$V$$, rather of difference in potential, $$\Delta V$$; this have to only it is in done once it is clear wherein the location of zero electric potential is defined.

We can define a free-falling mass by stating the the mass move from a region where it has actually high gravitational potential energy to a region of reduced gravitational potential power under the influence of the force of gravity (the force connected with a potential energy constantly acts in the direction to decreases potential energy). The exact same is true for electrical potential energy: charges will constantly experience a pressure in a direction to decrease their electric potential energy. However, optimistic charges will experience a pressure driving castle from regions of high electric potential to regions of low electric potential, whereas negative charges will endure a pressure driving them from areas of low electric potential to areas of greater electric potential. This is because, for negative charges, the adjust in potential energy associated with relocating through space, $$\Delta U$$, will be the an adverse of the corresponding readjust in electrical potential, $$\Delta U=q\Delta V$$, because the charge, $$q$$, is negative.

Exercise $$\PageIndex1$$

Electric potential boosts along the $$x$$ axis. A proton and also an electron are inserted at remainder at the origin; in i beg your pardon direction execute the charges move when released?

the proton moves towards negative $$x$$, when the electron move towards confident $$x$$. The proton move towards optimistic $$x$$, if the electron moves towards negative $$x$$. The proton and electron relocate towards negative $$x$$. The proton and also electron relocate towards hopeful $$x$$. Answer

If the only force exerted top top a fragment is the electric force, and also the fragment moves in an are such that the electrical potential changes by $$\Delta V$$, we have the right to use preservation of energy to identify the corresponding readjust in kinetic power of the particle:

\<\beginaligned \Delta E&=\Delta U+\Delta K=0 \\ \Delta U&=q\Delta V \endaligned\>

\<\therefore \Delta K=-q\Delta V\>

where $$\Delta E$$ is the adjust in full mechanical power of the particle, i beg your pardon is zero when power is conserved. The kinetic energy of a confident particle boosts if the fragment moves native a an ar of high potential come a region of short potential (as $$\Delta V$$ would certainly be an unfavorable and $$q$$ is positive), and also vice versa for a negative particle. This provides sense, because a positive and negative particle feel forces in the opposite directions.

In stimulate to describe the energies of particles such together electrons, that is convenient to usage a different unit of power than the Joule, so that the quantities involved are no orders of magnitude smaller sized than 1. A common an option is the “electron volt”, . One electron volt coincides to the power acquired through a fragment with a charge of $$e$$ (the fee of the electron) when it is accelerated by a potential difference of $$1\textV$$: \<\beginaligned \Delta E &= q\Delta V\\ 1\texteV&=(e)(1\textV)=1.6\times 10^-19\textJ\endaligned\> one electron that has sped up from rest throughout a an ar with a $$150\textV$$ potential difference across it will have a kinetic that $$150\texteV=2.4\times 10^-17\textJ$$. As you can see, that is less complicated to explain the energy of one electron in electron volts 보다 Joules.

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It is often beneficial in londonchinatown.orgics to take formerly learned concepts and compare lock to new ones, in this case, gravitational potential energy and also electric potential power can be compared to help understand the londonchinatown.orgical meaning of electrical potential.

Suppose that things with a big mass, $$M$$, is sitting in space. Currently place an object of a lot smaller mass, $$m$$, at any distance, $$r$$, indigenous the center of $$M$$. The gravitational potential energy of the tiny mass is offered by the complying with formula: \<\beginaligned U_g&=\fracGMmr\endaligned\> which is very comparable to the formula for electrical potential energy: \<\beginaligned U(\vec r)&=\frackQqr\endaligned\> Now, if us were to remove the fixed $$m$$ from its position, we would no much longer have an object with gravitational potential energy. However, we could still explain the gravitational potential because that the point, $$r$$, i m sorry would an outcome in gravitational potential energy when any type of mass $$m$$ is inserted there. This is the gravitational identical to electrical potential, and can be identified as: \<\beginaligned V_g&=\fracU_gm\endaligned\> which is also very comparable to the formula for electric potential: \<\beginaligned V_E&=\fracU_Eq\endaligned\> This to compare is shown in number $$\PageIndex1$$. Figure $$\PageIndex1$$: Gravitational potential energy and gravitational potential (left) alongside its electric analogue (right).

Example $$\PageIndex1$$

A proton and an electron move from a region of room where the electrical potential is $$20\textV$$ come a region of space where the electric potential is $$10\textV$$. If the electrical force is the only pressure exerted ~ above the particles, what can you say around their change in speed?

Solution:

The 2 particles relocate from a region of space where the electrical potential is $$20\textV$$ to a an ar of space where the electric potential is $$10\textV$$. The adjust in electrical potential experienced by the corpuscle is thus: \<\beginaligned \Delta V = V_final-V_initial=(10\textV)-(20\textV)=-10\textV\endaligned\> and we take the possibility to emphasize the one have to be an extremely careful with indications when utilizing potential. The adjust in potential energy of the proton, v charge $$q=+e$$, is thus: \<\beginaligned \Delta U_p=q\Delta V = (+e)(-10\textV)=-10\texteV\endaligned\> The potential power of the proton thus decreases by $$10\texteV$$ (which you can easily transform to Joules). Since we room told that no other pressure is exerted ~ above the particle, the total mechanical energy of the bit (kinetic to add potential energies) must be constant. Thus, if the potential power decreased, then the kinetic power of the proton has actually increased by the exact same amount, and the proton’s rate increases.

The readjust in potential power of the electron, with charge $$q=-e$$, is thus: \<\beginaligned \Delta U_e=q\Delta V = (-e)(-10\textV) = 10\texteV\endaligned\> The potential energy of the electron for this reason increases through $$10\texteV$$. Again, the mechanical power of the electron is conserved, so that boost in potential energy results in the same decrease in kinetic energy and also the electron’s speed decreases.

Discussion:

By utilizing the electric potential, $$V$$, us modelled the change in electric potential power of a proton and an electron together they both relocated from one region of an are to another.

We found that once a proton move from a an ar of high electric potential to a an ar of lower electric potential, that is potential energy decreases. This is since the proton has actually a optimistic charge and a to decrease in electrical potential will also result in a diminish in potential energy. Since no other forces are exerted ~ above the proton, the proton’s kinetic energy must increase. Since the potential power of the proton decreases, the proton is moving in the same direction together the electrical force, and the electric force go positive occupational on the proton to boost its kinetic energy.

Conversely, we found that when an electron moves from a an ar of high electrical potential to a an ar of lower electric potential, that is potential energy increases. This is due to the fact that it has actually a negative charge and also a diminish in electrical potential for this reason results in boost in potential energy. Because no other forces are exerted on the electron, the electron’s kinetic power must decrease, and also the electron slows down. This renders sense, because the force that is exerted on an electron will certainly be in the opposite direction from the pressure exerted ~ above a proton.

Exercise $$\PageIndex3$$

What reasons a positive charged particle to obtain speed as soon as it is accelerated through a potential difference?:

The particle increases because it loses potential power as it move from high to low potential. The particle speeds up because it loses potential power as it move from low to high potential The particle increases because it gains potential energy. The particle increases because it moves towards an adverse charges. Answer

Example $$\PageIndex2$$

What is the electrical potential at the leaf of a hydrogen atom (a distance of $$1$$ indigenous the proton), if one set $$0\text V$$ in ~ infinity? If an electron is located at a distance of $$1$$ from the proton, how much power is required to eliminate the electron; the is, exactly how much energy is required to ionize the hydrogen atom?

Solution:

We can easily calculate the electric potential, a distance of $$1\unicodexC5$$ indigenous a proton, because this corresponds to the potential from a suggest charge (with $$C=0$$): \<\beginaligned V(\vec r)=\frackQr=\frac(9\times 10^9\textN\cdot\textm^2\text/C^2)(1.6\times 10^-19\textC)(1\times 10^-10\textm)=14.4\textV\endaligned\> We deserve to calculate the potential energy of the electron (relative come infinity, where the potential is $$0\text V$$, since we chose $$C=0$$): \<\beginaligned U=(-e)V=(-1.6\times 10^-19\textC)(14.4\textV)=-14.4\texteV=-2.3\times 10^-18\textJ\endaligned\> whereby we also expressed the potential energy in electron volts. In stimulate to remove the electron from the hydrogen atom, we should exert a force (do work) until the electron is infinitely far from the proton. In ~ infinity, the potential energy of the electron will certainly be zero (by our selection of $$C=0$$). When relocating the electron native the hydrogen atom to an infinite distance away, we must do positive occupational to respond to the attractive pressure from the proton. The job-related that we must do is specifically equal come the change in potential power of the electron (and same to the an adverse of the work done by the pressure exerted by the proton):

\<\beginaligned W=\Delta U=(U_final-U_initial)=(0\textJ--2.3\times 10^-18\textJ)=2.3\times 10^-18\textJ \endaligned\>

The positive work that we must do, exerting a force that is opposite come the electric force, is positive and equal to $$2.3\times 10^-18\textJ$$, or $$14.4\texteV$$. If you look up the ionization energy of hydrogen, friend will find that it is $$13.6\texteV$$, so the this really simplistic design is quite specific (we could improve the design by adjusting the proton-electron distance so the the potential is $$13.6\textV$$).

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Discussion:

In this example, we established the electrical potential energy of one electron in a hydrogen atom, and found the it is negative, as soon as potential power is defined to it is in zero at infinity. In order to eliminate the electron from the atom, we must do positive work in bespeak to rise the potential energy of the electron from a negative value to zero (the potential energy at infinity). This is analogous to the work that should be done on a satellite in a gravitationally tied orbit because that it to reach escape velocity.