Themes > Science > Physics > Electromagnetism > Electrostatics > Electrostatic Potential > Electrostatic Potential

In general, if charges are in the presence of an electric field ${\bf E}$, they will acquire a potential energy. Recall that if the force is conservative, then, as discussed in Section 3.7, force can be replaced by potential energy, denoted by $U$. In the case of the Coulomb force, a direct demonstration can be made to show that it is a conservative force. Since force is proportional to the electric field, it follows that the (stationary) electric field can be expressed as the gradient of the ``electrostatic potential'', denoted by $V$.


% latex2html id marker 10601 \fbox{\fbox{\parbox{12cm}{ The relation between po... ...\Rightarrow -\nabla U&=&-q \nabla V\\ \Rightarrow U&=&q V \end{eqnarray} }}}


The force on a charge $q$ due to an electric field ${\bf E}$ is given by

\begin{displaymath} {\bf F}=q{\bf E} \end{displaymath} (7.9)

and hence
\begin{displaymath} \Rightarrow U=q V \end{displaymath} (7.10)

In other words, the potential energy $U$ is proportional to the electrostatic potential $V$. Consider the two points $d_1,d_2$ with electrostatic potential $V_1,V_2$; the electric field is then given by
$\displaystyle E$ $\textstyle =$ $\displaystyle -\frac{V_2-V_1}{d_2-d_1}$ (7.11)
  $\textstyle =$ $\displaystyle -\frac{\Delta V}{\Delta d}$ (7.12)

The $E$ field is the negative of the gradient of the electrostatic potential so that it points from a higher value of $V$ to a lower value, in keeping with the force pointing in the direction of decreasing potential.

 

Figure 7.7: Electric Field from Electrostatic Potential
\begin{figure} \begin{center} \epsfig{file=core/potential1.eps, width=10cm} \end{center} \end{figure}

 

Since $[U]=ML^2T^{-2}$, the dimensions of the electrostatic potential is given by $[V]=ML^2T^{-2}C^{-1}$. The units for $V$ in the SI system is called Volts (V) and is given by
\begin{displaymath} V=\mathrm{Joule/Coulomb}\equiv JC^{-1} \end{displaymath} (7.13)

In terms of volts, we have a simpler expression for the units of electric field. Recall that $[{\bf E}]=NC^{-1}$. In term of volts, we have
\begin{displaymath}[{\bf E}]=\mbox{\rm {volts per meter}}\equiv VM^{-1} \end{displaymath} (7.14)

Note that the electrostatic potential $V$ is not a vector; rather at every point in space $x$ only a single number $V(x)$ completely specifies the $V$. A field such as $V(x)$ is called a scalar field. Clearly it is much easier to analyze a scalar field such as $V$ compared to a vector field like ${\bf E}$, and this is one of the main reasons for working with potentials. The contours of constant $V$ lines are plotted in Figure 7.7. Since the electrostatic force is conservative, the energy of a particle at some point is independent of the path it took to get there. As shown in Figure 7.7, the difference in the energy of a charged particle is the same whether it takes path 1 or path 2 in going from the surface $V=5V$ to the surface $V=6V$. Figure 7.8 shows the lines of equipotential in three dimensions. The lines with a high value of $V$ are analogous to a point of high elevation for a body moving under the influence of gravity. Just as in gravity a mass $m$ can gain potential energy in going from a lower to a higher height, a positively charged particle also gains potential energy in moving from a point with a lower to a point with a higher electrostatic potential $V$; however, and this is what makes electromagnetism so different from gravity, negatively charged particle loses energy in moving from a point with a lower value of $V$ to one with a higher value.

 

Figure 7.8: Equipotential Surface in Three Dimensions
\begin{figure} \begin{center} \epsfig{file=core/Equipotential1.eps, width=6cm} \end{center} \end{figure}


A charged particle will take the shortest path in going from a point at a higher to a point at a lower potential. In effect, the charge moves under the force created by the electric field derived from the potential $V$. The direction of the electric field at some point on a contour points in the direction of the shortest distance to close-by contour points, as shown in Figure 7.7. The magnitude of ${\bf E}$ is determined by the steepness of the contours. What is the rationale for separating out the electrostatic potential from the potential energy of two charges? In contrast to say gravity, for which mass is always a positive quantity, electric charge can either be negative or positive. Hence, by separating off the electrostatic potential $V$ from potential $U$, we can study the field $V$ generated by a given charge, and then analyze how it affects other charges. Suppose $V$ is generated by a positive charge; a negative charge will move towards increasing values of $V$ whereas a positive charge will move towards decreasing values of $V$, as shown in Figure 7.9. A potential $V$ that looks like a ``mountain'' to a positively charged particle $+q$ looks like a ``crater'' to a negatively charged particle $-q$.
Figure 7.9: Positive and Negative Charge Moving in the Same V-field
\begin{figure} \begin{center} \epsfig{file=core/potential2.eps, width=6cm} \end{center} \end{figure}
Recall that the fundamental irreducible charge in nature is the electron charge; by convention, the charge of the electron is taken to be negative and is denoted by $-e$. It is a natural constant, with the numerical value of $e$ in the SI units given by
\begin{displaymath} e=1.60\times 10^{-19} C \end{displaymath} (7.15)

The value of the electron charge is extremely small. In a typical light bulb, every second over $10^{19}$ electron charges enter and leave the bulb's filament. Protons carry charge equal to $+e$, and the proton seldom enters the processes that we will be interested in. The electron charge is the basis of another unit for energy. The energy gained by an electron, in moving from say a point with $V=5$ volts in Figure 7.7 to a point with $V=6$ volts, such that the electrostatic potential difference between the two points is 1 volt is called an electron-volt, and is denoted by $eV$. We consequently have
$\displaystyle 1 eV$ $\textstyle =$ $\displaystyle e \times 1 \mathrm{volt}$ (7.16)
  $\textstyle =$ $\displaystyle 1.60 C\times 10^{-19} \times JC^{-1}$ (7.17)
$\displaystyle \Rightarrow 1 eV$ $\textstyle =$ $\displaystyle 1.60\times 10^{-19}J$ (7.18)


The electron-volt (eV) is the appropriate unit of energy to measure energies involved in atomic, chemical and (molecular) biological processes. The energy of a single molecule of air at room temperature is about $\displaystyle \frac{1}{40}$ eV.

Potential due to Point Charges

The potential energy of charges $q_1$ and $q_2$, separated by a distance $r$, is given by Coulomb potential
\begin{displaymath} U_{\mathrm{Coulomb}}(r)=k\frac{q_1q_2}{r} \mbox{\rm { : Electrostatic Potential Energy}} \end{displaymath} (7.19)

From eq.(7.17) the electrostatic potential of a point charge $q_2$ is given by
$\displaystyle V(r)$ $\textstyle =$ $\displaystyle \frac{U_{\mathrm{Coulomb}}(r)}{q_1}$ (7.20)
$\displaystyle \Rightarrow V(r)$ $\textstyle =$ $\displaystyle k\frac{q_2}{r}$ (7.21)

Figure 7.10 shows the Coulomb potential. Note the striking difference in the potentials due to a positive and negative charge, with the positive charge giving a ``mountain'' and the negative charge yielding a ``valley''.

 

Figure 7.10: Coulomb Potential for Positive and Negative Charge
\begin{figure} \begin{center} \epsfig{file=core/potential.eps, width=10cm} \end{center} \end{figure}


Information provided by: http://www.srikant.org