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Next: Continuous distribution of charge, Up: Electrostatics Previous: Electrostatics

Point Charges

Two charges $e_{1}$ and $e_{2}$ ($coulombs$) a distance $r$ ($m$) apart in a vacuum feel a force ($newtons$)
\begin{displaymath}
{e_{1}e_{2}\over 4 \pi \epsilon_{0}r^{2}}.
\end{displaymath} (3.5)

The factor $4\pi$ simplifies later formulae. If $e_{1}$ and $e_{2}$ are of the same sign then the force is positive and so is repulsive. If they are of opposite signs the force is negative and so is attractive. Calculating the magnitude of the coefficients we find that

\begin{displaymath}
{1\over 4\pi \epsilon_{0}} \approx 9 \times 10^{9},
\end{displaymath}

so that the Coulomb force is enormous. This means that ordinary matter is essentially neutral.

If a unit charge is placed at a point $P$, the force it experiences is called the electric field at $P$. A point charge, $e$, at $O$ produces at $P$ an electric field (i.e. the force a unit charge feels there),

\begin{displaymath}
{\bf E} = {e\over 4\pi \epsilon_{0} r^{2}}{\bf e}_{r}.
\end{displaymath} (3.6)

In electrostatics $\partial /\partial t = 0$ and so (3.2) becomes

\begin{displaymath}
\nabla \times {\bf E} = 0,
\end{displaymath}

which is satisfied by setting
\begin{displaymath}
{\bf E} = -\nabla F.
\end{displaymath} (3.7)

$F$ is called the electrostatic potential. For the point charge situation described above this is
\begin{displaymath}
F = {e\over 4\pi \epsilon_{0}r}.
\end{displaymath} (3.8)

Note that :
  1. The potential due to several charges is obtained by adding the individual potentials together so that

    \begin{displaymath}
F = {1\over 4\pi \epsilon_{0}}\left ({e_{1}\over r_{1}} +
{e_{2}\over
r_{2}} + \cdots \right ).
\end{displaymath}

  2. In general, surfaces of constant $F$ are perpendicular to the electric field ${\bf E}$.

Example 3. .1Consider an electron of mass $m$ and charge $-e$ moving in a circlar orbit of radius $r$ about a fixed charge $+e$ at the origin $O$. Obtain the speed of the electon.

The electron feels a centrifugal force of $m v^{2}/r$, due to the circular motion and this is balanced by the attractive Coulomb force of $e^{2}/4\pi \epsilon_{0}
r^{2}$. Therefore,

\begin{displaymath}
m{v^{2}\over r} = {e^{2}\over 4 \pi \epsilon_{0}r^{2}}.
\end{displaymath}

Hence, the electron's speed is

\begin{displaymath}
v = \sqrt{{e^{2}\over 4 \pi \epsilon_{0}rm}}
\end{displaymath}

Example 3. .2Two positive charges of strength $e$ are located at $(0,0,\pm a)$ on the z-axis. Find the electric field on the z-axis and sketch the field lines in the x-z plane.

Figure 3.1: Two point charges of strength $e$ are located at (0,0,-a) and (0,0,a).
\includegraphics [scale=1.0]{fundfig31.eps}

The position vector of a general point in the $x-z$ plane is

\begin{displaymath}
{\bf r} = x{\bf i} + z{\bf k}.
\end{displaymath}

The situation is shown in Figure 3.1. On the z-axis ($x=0$), we need to consider the direction of the electric field due to each of the point charges. Starting to the left of $z=-a$, both point charges give a component of the electric field that is in the negative ${\bf k}$ direction. Between the point charges the left hand charge gives a field component that points to the right and the right hand charge points to the left. Finally, to the right of the charges, both give electric field components pointing to the right. Thus, we have

\begin{eqnarray*}
z < -a, \qquad 4\pi \epsilon_{0}E & = & -{e\over (z+a)^{2}} -...
...i \epsilon_{0}E & = & {e\over (z+a)^{2}} +
{e\over (z-a)^{2}}.
\end{eqnarray*}



To calculate the electric field at a general point we need to work out the distance from each point charge and the unit vector. For example,

\begin{displaymath}
{\bf r_{1}} = {\bf r} + {\bf a} = x{\bf i} + (z + a){\bf k},
\end{displaymath}

and

\begin{displaymath}
{\bf r_{2}} = {\bf r} - {\bf a} = x{\bf i} + (z - a){\bf k}.
\end{displaymath}

The unit vectors are

\begin{displaymath}
{\hat{\bf r_{1}}} = {{\bf r_{1}}\over (x^{2} +
(z+a)^{2})...
...at{\bf r_{2}}} = {{\bf r_{2}}\over (x^{2} + (z-a)^{2})^{1/2}}.
\end{displaymath}

Using these results, the electric field is given by

\begin{displaymath}
{\bf E} = {e\over 4\pi \epsilon_{0}}{x{\bf i} + (z+a){\bf k...
...n_{0}}{x{\bf i} + (z-a){\bf k}\over (x^{2} + (z-a)^{2})^{3/2}}
\end{displaymath}

The resulting electric field is shown in Figure 3.2

Figure 3.2: The electric field with a netural point at $N$.
\includegraphics [scale=0.7]{fundfig32.eps}

Example 3. .3Charges $2e$ and $-e$ are located on the x-axis at (-a,0,0) and (a,0,0) respectively. (i) Show that there is one neutral point, $N$, where the electric field vanishes, i.e.

\begin{displaymath}
{\bf E} = 0 \hbox{ at a Neutral Point}.
\end{displaymath}

(ii) Draw a rough sketch of the electric field lines.

(i) Consider the general point, $P$, ${\bf r} = x{\bf i} + z{\bf k}$. The situation is shown in Figure 3.3.

Figure 3.3: A point charge of strength $2e$ is located at $(-a,0,0)$ and a negative point charge $-e$ is located at $( a,0, 0)$.
\includegraphics [scale=0.7]{fundfig32a.eps}

The electric fields due to each of the point charges are only colinear (and therefore can cancel at a neutral point) when $P$ is on the x-axis. Here they cancel at a point $N$. Consider the electric field on the x-axis. As in the previous example, the form of the electric field depends on where the point is. Thus,

\begin{eqnarray*}
x <-a, \qquad 4 \pi \epsilon_{0}E & = & -{2e\over (x+a)^{2}} ...
... \epsilon_{0}E & = & {2e\over (x+a)^{2}} -
{e\over (x-a)^{2}}.
\end{eqnarray*}



Thus, there is only a possible neutral point in the first and third regions. To obtain the position of the neutral point we solve

\begin{displaymath}
{2e\over (x+a)^{2}} = {e\over (x-a)^{2}}.
\end{displaymath}

Rearranging we obtain

\begin{eqnarray*}
2(x_{N}-a)^{2} & = & (x_{N}+a)^{2}, \\
\Rightarrow & & x_{N...
...w & & x_{N} = (3 + \sqrt{8})a, \qquad x_{N} = (3 -
\sqrt{8})a.
\end{eqnarray*}



Only the first of these roots lies in the correct range for $x_{N}$. The second root lies in the interval $0 < x_{N} < a$ where it is clear that there can be no neutral point. Thus, the only neutral point is at $x_{N} = (3 + \sqrt{8})a$.

(ii) To plot the field lines we can use the electrostatic potential where ${\bf E} = - \nabla F$. We can verify that

\begin{displaymath}
4\pi \epsilon_{0} F = {2e\over ((x+a)^{2} + z^{2})^{1/2}} - {e
\over ((x-a)^{2} + z^{2})^{1/2}}.
\end{displaymath}

The electric field lines are given by contours of constant $F$. Thus, we obtain the field lines as shown in Figure 3.4.

Figure 3.4: Electric field lines.
\includegraphics [scale=0.7]{fundfig33.eps}

Half the flux from the point source $2e$ goes into $-e$ and half goes off to infinity.
next up previous
Next: Continuous distribution of charge, Up: Electrostatics Previous: Electrostatics
Prof. Alan Hood
2000-11-06