# 18.5 Electric field lines: multiple charges  (Page 2/7)

 Page 2 / 7

Solution

The electric field strength at the origin due to ${q}_{1}$ is labeled ${E}_{1}$ and is calculated:

$\begin{array}{}{E}_{1}=k\frac{{q}_{1}}{{r}_{1}^{2}}=\left(8\text{.}\text{99}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{N}\cdot {\text{m}}^{2}{\text{/C}}^{2}\right)\frac{\left(5\text{.}\text{00}×{\text{10}}^{-9}\phantom{\rule{0.25em}{0ex}}\text{C}\right)}{{\left(2\text{.}\text{00}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m}\right)}^{2}}\\ {E}_{1}=1\text{.}\text{124}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}.\end{array}$

Similarly, ${E}_{2}$ is

$\begin{array}{}{E}_{2}=k\frac{{q}_{2}}{{r}_{2}^{2}}=\left(8\text{.}\text{99}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{N}\cdot {\text{m}}^{2}{\text{/C}}^{2}\right)\frac{\left(\text{10}\text{.}0×{\text{10}}^{-9}\phantom{\rule{0.25em}{0ex}}\text{C}\right)}{{\left(4\text{.}\text{00}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m}\right)}^{2}}\\ {E}_{2}=0\text{.}\text{5619}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}.\end{array}$

Four digits have been retained in this solution to illustrate that ${E}_{1}$ is exactly twice the magnitude of ${E}_{2}$ . Now arrows are drawn to represent the magnitudes and directions of ${\mathbf{\text{E}}}_{1}$ and ${\mathbf{\text{E}}}_{2}$ . (See [link] .) The direction of the electric field is that of the force on a positive charge so both arrows point directly away from the positive charges that create them. The arrow for ${\mathbf{\text{E}}}_{1}$ is exactly twice the length of that for ${\mathbf{\text{E}}}_{2}$ . The arrows form a right triangle in this case and can be added using the Pythagorean theorem. The magnitude of the total field ${E}_{\text{tot}}$ is

$\begin{array}{lll}{E}_{\text{tot}}& =& \left({E}_{1}^{2}+{E}_{2}^{2}{\right)}^{\text{1/2}}\\ & =& {\left\{\left(\text{1.124}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}{\right)}^{2}+\left(\text{0.5619}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}{\right)}^{2}\right\}}^{\text{1/2}}\\ & =& \text{1.26}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C.}\end{array}$

The direction is

$\begin{array}{lll}\theta & =& {\text{tan}}^{-1}\left(\frac{{E}_{1}}{{E}_{2}}\right)\\ & =& {\text{tan}}^{-1}\left(\frac{1\text{.}\text{124}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}}{0\text{.}\text{5619}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C}}\right)\\ & =& \text{63}\text{.}4º,\end{array}$

or $63.4º$ above the x -axis.

Discussion

In cases where the electric field vectors to be added are not perpendicular, vector components or graphical techniques can be used. The total electric field found in this example is the total electric field at only one point in space. To find the total electric field due to these two charges over an entire region, the same technique must be repeated for each point in the region. This impossibly lengthy task (there are an infinite number of points in space) can be avoided by calculating the total field at representative points and using some of the unifying features noted next.

[link] shows how the electric field from two point charges can be drawn by finding the total field at representative points and drawing electric field lines consistent with those points. While the electric fields from multiple charges are more complex than those of single charges, some simple features are easily noticed.

For example, the field is weaker between like charges, as shown by the lines being farther apart in that region. (This is because the fields from each charge exert opposing forces on any charge placed between them.) (See [link] and [link] (a).) Furthermore, at a great distance from two like charges, the field becomes identical to the field from a single, larger charge.

[link] (b) shows the electric field of two unlike charges. The field is stronger between the charges. In that region, the fields from each charge are in the same direction, and so their strengths add. The field of two unlike charges is weak at large distances, because the fields of the individual charges are in opposite directions and so their strengths subtract. At very large distances, the field of two unlike charges looks like that of a smaller single charge.

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