As you respeak to, we characterized the **electric field**, (vec E(vec r)), to be the **electrical pressure per unit charge**. By specifying an electric area anywhere in room, we were able to easily identify the force on any kind of test charge, (q), whether the test charge is positive or negative (because the authorize of (q) will adjust the direction of the force vector, (qvec E)): <eginaligned vec E(vec r) &= fracvec F^E(vec r)q\ herefore vec F^E(vec r)&=qvec E(vec r)endaligned> Similarly, we define the **electric potential**, (V(vec r)), to be the **electric potential energy per unit charge**. This enables us to define electrical potential, (V(vec r)), everywhere in area, and then identify the potential energy of a particular charge, (q), by simply multiplying (q) through the electric potential at that position in area. <eginaligned V(vec r) &= frac U(vec r)q\ herefore U(vec r)&= q V(vec r)endaligned> The S.I. unit for electric potential is the “volt”, (V). Electric potential, (V(vec r)), is a scalar field whose worth is “the electric potential” at that position in space. A positive charge, (q=1 extC), will certainly thus have a potential energy of (U=10 extJ) if it is situated at a position in area wbelow the electric potential is (V=10 extV), considering that (U=qV). Similarly, a negative charge, (q=-1 extC), will have negative potential power, (U=-10 extJ), at the exact same place.

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Due to the fact that only distinctions in potential energy are slrfc.orgically coherent (as change in potential energy is regarded work), **just alters in electric potential are slrfc.orgically meaningful** (as electrical potential is related to electric potential energy). A difference in electric potential is typically called a “voltage”. One often provides a clear choice of wright here the electrical potential is zero (frequently the ground, or infinitely much away), so that the term voltage is offered to describe potential, (V), rather of distinction in potential, (Delta V); this must only be done as soon as it is clear wbelow the place of zero electric potential is defined.

We deserve to define a free-falling mass by stating that the mass moves from a region where it has high gravitational potential power to an area of lower gravitational potential power under the affect of the force of gravity (the force linked with a potential energy constantly acts in the direction to decreases potential energy). The same is true for electric potential energy: **charges will always suffer a pressure in a direction to decrease their electrical potential energy**. However before, positive charges will experience a force driving them from regions of high electrical potential to areas of low electrical potential, whereas negative charges will experience a force driving them from areas of low electrical potential to areas of greater electric potential. This is bereason, for negative charges, the readjust in potential energy connected through moving via room, (Delta U), will certainly be the negative of the corresponding readjust in electric potential, (Delta U=qDelta V), given that the charge, (q), is negative.

Exercise (PageIndex1)

Electric potential increases alengthy the (x) axis. A proton and an electron are placed at remainder at the origin; in which direction do the charges move when released?

the proton moves in the direction of negative (x), while the electron moves towards positive (x). the proton moves towards positive (x), while the electron moves in the direction of negative (x). the proton and also electron relocate in the direction of negative (x). the proton and also electron move towards positive (x).**Answer**

If the only force exerted on a particle is the electric force, and also the pwrite-up moves in room such that the electric potential alters by (Delta V), we can use conservation of energy to recognize the equivalent adjust in kinetic energy of the particle:

<eginaligned Delta E&=Delta U+Delta K=0 \ Delta U&=qDelta V endaligned>

< herefore Delta K=-qDelta V>

where (Delta E) is the change in total mechanical power of the pwrite-up, which is zero when power is conserved. The kinetic energy of a positive particle rises if the pwrite-up moves from an area of high potential to a region of low potential (as (Delta V) would be negative and (q) is positive), and also vice versa for a negative particle. This provides sense, considering that a positive and negative pwrite-up feel forces in oppowebsite directions.

In order to describe the energies of pshort articles such as electrons, it is convenient to usage a different unit of energy than the Joule, so that the amounts affiliated are not orders of magnitude smaller than 1. A common option is the “electron volt”, . One electron volt corresponds to the power gained by a ppost with a charge of (e) (the charge of the electron) as soon as it is increased by a potential difference of (1 extV): <eginaligned Delta E &= qDelta V\ 1 exteV&=(e)(1 extV)=1.6 imes 10^-19 extJendaligned> An electron that has actually sped up from rest throughout a region via a (150 extV) potential distinction across it will certainly have actually a kinetic of (150 exteV=2.4 imes 10^-17 extJ). As you deserve to see, it is easier to describe the power of an electron in electron volts than Joules.

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It is frequently useful in slrfc.orgics to take formerly learned principles and also compare them to new ones, in this instance, gravitational potential power and electric potential power deserve to be compared to aid understand the slrfc.orgical meaning of electric potential.

Suppose that a things with a large mass, (M), is sitting in space. Now location an object of a much smaller sized mass, (m), at any distance, (r), from the facility of (M). The gravitational potential power of the little mass is offered by the following formula: <eginaligned U_g&=fracGMmrendaligned> Which is extremely comparable to the formula for electric potential energy: <eginaligned U(vec r)&=frackQqrendaligned> Now, if we were to rerelocate the mass (m) from its position, we would no longer have actually an object via gravitational potential power. However, we might still explain the gravitational potential for the point, (r), which would certainly bring about gravitational potential power once any kind of mass (m) is put tright here. This is the gravitational tantamount to electrical potential, and also deserve to be defined as: <eginaligned V_g&=fracU_gmendaligned> which is also incredibly equivalent to the formula for electric potential: <eginaligned V_E&=fracU_Eqendaligned> This comparison is illustrated in Figure (PageIndex1).

Figure (PageIndex1): Gravitational potential energy and also gravitational potential (left) alongside its electric analogue (right).Example (PageIndex1)

A proton and also an electron move from a region of space where the electrical potential is (20 extV) to a region of space wbelow the electric potential is (10 extV). If the electrical force is the just force exerted on the pwrite-ups, what can you say around their change in speed?

**Solution**:

The two pwrite-ups move from an area of room where the electrical potential is (20 extV) to a region of space where the electrical potential is (10 extV). The readjust in electric potential experienced by the pwrite-ups is thus: <eginaligned Delta V = V_final-V_initial=(10 extV)-(20 extV)=-10 extVendaligned> and we take the chance to emphasize that one need to be extremely cautious through indications as soon as using potential. The adjust in potential energy of the proton, through charge (q=+e), is thus: <eginaligned Delta U_p=qDelta V = (+e)(-10 extV)=-10 exteVendaligned> The potential power of the proton thus decreases by (10 exteV) (which you deserve to quickly transform to Joules). Because we are told that no various other pressure is exerted on the ppost, the full mechanical power of the pshort article (kinetic plus potential energies) need to be continuous. Thus, if the potential power lessened, then the kinetic power of the proton has actually raised by the exact same amount, and also **the proton’s rate increases**.

The change in potential power of the electron, through charge (q=-e), is thus: <eginaligned Delta U_e=qDelta V = (-e)(-10 extV) = 10 exteVendaligned> The potential power of the electron for this reason increases by (10 exteV). Aget, the mechanical energy of the electron is conserved, so that an increase in potential energy results in the very same decrease in kinetic energy and **the electron’s rate decreases**.

**Discussion:**

By utilizing the electric potential, (V), we modelled the change in electric potential energy of a proton and also an electron as they both moved from one area of space to another.

We found that as soon as a **proton moves from a region of high electrical potential to a region of reduced electric potential, its potential energy decreases**. This is because the proton has a positive charge and also a decrease in electrical potential will certainly additionally result in a decrease in potential power. Due to the fact that no various other pressures are exerted on the proton, the proton’s kinetic energy have to rise. Because the potential energy of the proton decreases, the proton is moving in the very same direction as the electric force, and also the electrical force does positive work on the proton to boost its kinetic power.

Conversely, we discovered that as soon as an **electron moves from a region of high electric potential to a region of lower electric potential, its potential energy increases**. This is because it has an adverse charge and also a decrease in electric potential for this reason results in a boost in potential energy. Because no various other forces are exerted on the electron, the electron’s kinetic power need to decrease, and also the electron slows down. This makes feeling, because the force that is exerted on an electron will be in the opposite direction from the pressure exerted on a proton.

Exercise (PageIndex3)

What reasons a positively charged pwrite-up to obtain speed as soon as it is sped up with a potential difference?:

The pwrite-up accelerates bereason it loses potential power as it moves from high to low potential. The particle increases because it loses potential energy as it moves from low to high potential The particle accelerates because it gains potential power. The pwrite-up accelerates bereason it moves in the direction of negative charges.**Answer**

Example (PageIndex2)

What is the electrical potential at the edge of a hydrogen atom (a distance of (1) from the proton), if one sets (0 ext V) at infinity? If an electron is situated at a distance of (1) from the proton, how a lot energy is compelled to rerelocate the electron; that is, just how much energy is forced to ionize the hydrogen atom?

**Solution**:

We can easily calculate the electrical potential, a distance of (1unicodexC5) from a proton, considering that this synchronizes to the potential from a suggest charge (with (C=0)): <eginaligned V(vec r)=frackQr=frac(9 imes 10^9 extNcdot extm^2 ext/C^2)(1.6 imes 10^-19 extC)(1 imes 10^-10 extm)=14.4 extVendaligned> We have the right to calculate the potential power of the electron (family member to infinity, where the potential is (0 ext V), considering that we chose (C=0)): <eginaligned U=(-e)V=(-1.6 imes 10^-19 extC)(14.4 extV)=-14.4 exteV=-2.3 imes 10^-18 extJendaligned> where we also expressed the potential energy in electron volts. In order to remove the electron from the hydrogen atom, we must exert a force (execute work) until the electron is infinitely far from the proton. At infinity, the potential energy of the electron will be zero (by our alternative of (C=0)). When relocating the electron from the hydrogen atom to an limitless distance away, we should carry out positive work to counter the attrenergetic force from the proton. The occupational that we should perform is exactly equal to the change in potential power of the electron (and equal to the negative of the occupational done by the pressure exerted by the proton):

<eginaligned W=Delta U=(U_final-U_initial)=(0 extJ--2.3 imes 10^-18 extJ)=2.3 imes 10^-18 extJ endaligned>

The positive job-related that we must do, exerting a pressure that is opposite to the electrical pressure, is positive and also equal to (2.3 imes 10^-18 extJ), or (14.4 exteV). If you look up the ionization energy of hydrogen, you will certainly uncover that it is (13.6 exteV), so that this extremely simplistic model is fairly precise (we can boost the model by adjusting the proton-electron distance so that the potential is (13.6 extV)).

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**Discussion:**

In this instance, we established the electric potential energy of an electron in a hydrogen atom, and found that it is negative, when potential power is defined to be zero at infinity. In order to rerelocate the electron from the atom, we need to perform positive job-related in order to increase the potential power of the electron from a negative worth to zero (the potential power at infinity). This is analogous to the work-related that need to be done on a satellite in a gravitationally bound orlittle for it to reach escape velocity.