In the previous section, we looked at the potential energy $U$ associated with a test charge in an electric field. Now we want to describe this potential energy on a “per unit charge” basis, just as electric field describes the force per unit charge on a charged particle in the field. This leads us to the concept of electric potential, often called simply potential. This concept is very useful in calculations involving energies of charged particles. It also facilitates many electric-field calculations because electric potential is closely related to the electric field $\mathbf{E}$. When we need to determine an electric field, it is often easier to determine the potential first and then find the field from it.
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| Fig.1:The voltage of this battery equals the difference in potential $V_{ab}=V_a-V_b$ between its positive terminal (point a) and its negative terminal (point b). |
$V=\frac{U}{q_0}$ or $U=q_0V$ (1)
Potential energy and charge are both scalars, so potential is a scalar. From Eq. (23.12) its units are the units of energy divided by those of charge. The SI unit of potential, called one volt (1 V) in honor of the Italian electrical experimenter Alessandro Volta (1745–1827), equals 1 joule per coulomb:
1 V = 1 volt = 1 J/C = 1 joule/coulomb
Let’s put Eq. (23.2), which equates the work done by the electric force during a displacement from a to b to the quantity $-\Delta U=-(U_b-U_a)$ on a “work per unit charge” basis. We divide this equation by $q_0$ obtaining
$\frac{W_{a→b}}{q_0}=-\frac{\Delta U}{q_0}=-\left(\frac{U_b}{q_0}-\frac{U_a}{q_0}\right)$
$\frac{W_{a→b}}{q_0}=-(V_b-V_a)=V_a-V_b$ (2)
where $V_a=U_a/q_0$ is the potential energy per unit charge at point a and similarly for $V_b$. We call $V_a$ and $V_b$ the potential at point a and potential at point respectively. Thus the work done per unit charge by the electric force when a charged body moves from a to b is equal to the potential at a minus the potential at b.
The difference $V_a-V_b$ is called the potential of a with respect to b, we sometimes abbreviate this difference as $V_{ab}=V_a-V_b$ (note the order of the subscripts). This is often called the potential difference between a and b, but that’s ambiguous unless we specify which is the reference point. In electric circuits, which we will analyze in later chapters, the potential difference between two points is often called voltage (Fig. 23.11). Equation (23.13) then states: $\mathbf{V_{ab}}$ the potential of a with respect to b equals the work done by the electric force when a UNIT charge moves from a to b.
Another way to interpret the potential difference $V_{ab}$ in Eq. (23.13) is to use the alternative viewpoint, $U_a-U_b$ mentioned at the end of Section 23.1. In that viewpoint, is the amount of work that must be done by an external force to move a particle of charge $q_0$ slowly from b to a against the electric force. The work that must be done per unit charge by the external force is then $(U_a-U_b)/q_0=V_a-V_b=V_{ab}$. In other words: $\mathbf{V_{ab}}$ the potential of a with respect to b, equals the work that must be done to move a UNIT charge slowly from b to a against the electric force.
POTENTIAL DUE TO A POINT CHARGE
To find the potential V due to a single point charge we divide Eq.(1) by $q_0$:
$V=\frac{U}{q_0}=\frac{1}{4\pi \epsilon_0}\frac{q}{r}$ (potential due to a point charge) (3)
where r is the distance from the point charge q to the point at which the potential is evaluated. If q is positive, the potential that it produces is positive at all points; if q is negative, it produces a potential that is negative everywhere. In either case, V is equal to zero at $r→\infty$, an infinite distance from the point charge. Note that potential, like electric field, is independent of the test charge $q_0$ that we use to define it.
Equation (2.8) is true for any sign of the charge Q, though we considered Q > 0 in its derivation. For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point P is positive. [This is as it should be, since for Q < 0, the force on a unit positive test charge is attractive, so that the electrostatic force and the displacement (from infinity to P) are in the same direction.] Finally, we note that Eq. (2.8) is consistent with the choice that potential at infinity be zero.
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| Fig.2: Variation of potential V with r [in units of ($Q/4\pi \epsilon_0) \ m^{-1}$](blue curve) and field with r [in units of ($Q/4\pi \epsilon_0) \ m^{-2}$] (black curve) for a point charge Q. |
Figure (2) shows how the electrostatic potential ($\propto 1/r$) and the electrostatic field ($\propto 1/r^2$) varies with r.
Example 1
(a) Calculate the potential at a point P due to a charge of $4 \times 10^{–7} \ C$ located 9 cm away. (b) Hence obtain the work done in bringing a charge of $2 \times 10^{–9} \ C$ from infinity to the point P. Does the answer depend on the path along which the charge is brought?
Solution:
(a) $V=\frac{1}{4\pi \epsilon_0}\frac{Q}{r}$
$V=(9 \times 10^9 N.m^2/C^2)\frac{4 \times 10^{–7} \ C }{0.09 m}=4\times 10^4 \ V$
(b) $W = qV = (2 \times 10^{–9} \ C)(4\times 10^4 \ V)$
$W=8\times 10^{-4} \ V$
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r and another perpendicular to r. The work done corresponding to the later will be zero.
Example 2
A proton (charge $+e=1.602 \times 10^{-19} \ C$) moves a distance $d=0.50 \ m$ in a straight line between points a and b in a linear accelerator. The electric field is uniform along this line, with magnitude $E = 1.5 \times 10^7 \ V/m$ in the direction from a to b. Determine (a) the force on the proton; (b) the work done on it by the field; (c) the potential difference $V_a - V_b$.
Solution:
This problem uses the relationship between electric field and electric force. It also uses the relationship among force, work, and potential-energy difference. We are given the electric field, so it is straightforward to find the electric force on the proton. Calculating the work is also straightforward because is uniform, so the force on the proton is constant. Once the work is known, we find using Eq. (23.13).
Solution:
(a) The force on the proton is in the same direction as the electric field, and its magnitude is
$F=qE=(1.602 \times 10^{-19} \ C)(1.5 \times 10^7 \ V/m)$
$F=2.4 \times 10^{-12} \ N$
(b) The force is constant and in the same direction as the displacement, so the work done on the proton is
$W_{a→b}=Fd = (2.4 \times 10^{-12} \ N)(0.50 \ m)$
$=1.2 \times 10^{-12} \ N$
$=(1.2 \times 10^{-12} \ N)\frac{1 \ eV}{(1.602 \times 10^{-19} \ C)}$
$W_{a→b}=7.5 \times 10^{6} \ eV=7.5 \ MeV$
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