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Reading: Chapter 21, ELECTRIC CHARGE AND ELECTRIC FIELD
Study guide: Chapter 22
A rubbed (statically charged) balloon sticks to a wall
(which hasn't been rubbed!)
We understand this in terms of an induced charge [PhET].
Electric force $$F_e = k \frac{q_1\cdot q_2}{d^2}$$
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Gravitational force
$$F_g=G \frac{m_1\cdot m_2}{d^2}$$
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With a "source" charge $q_1$ at the origin, and a "test" charge at position $\myv r$, the force that the test charge $q_2$ feels is:
$$\myv F_e= k\frac{q_1q_2}{r^2}\uv r.$$
$$\myv F_{e1} = \left(k\frac{q_1}{r^2}\uv r\right) q = \myv E_1 q$$
So
$$\myv E_1 = \frac{\myv F_{e1}}{q_1}$$
The electrical field at a position $\myv r$ points in the same direction as the force that a positive charge at that position would feel.
$$\begineq \myv F_\text{net} &=& \myv F_1 + \myv F_2 +\myv F_3+...\\
&=& q (\myv E_1 + \myv E_2 +\myv E_3+...) \\
&=& q\myv E_\text{net}.\endeq$$
Textbooks use as unit vectors $\bf{i}\equiv \uv x$ and $\bf{j}\equiv \uv y$ and $\bf{k}\equiv \uv z$. In the figure,
$$\begineq \myv E_1&=& 6\times 10^6 N/C \,{\bf i}+3\times 10^6 N/C\,{\bf j};\\ \myv E_2&=& 4\times 10^6 N/C \,{\bf i}-6\times 10^6 N/C\,{\bf j}.\endeq$$
The net electric field $\myv E_\text{net}$ at the blue position can be calculated from the vector components:
$$\begineq\myv E_\text{net} &=& \left[(6+4){\bf i} +(3-6){\bf j}\right]\times 10^6 N/C\\ &=& \left[10\,{\bf i} -3\,{\bf j}\right]\times 10^6 N/C.\endeq$$
Ejiri in the Suruga Province, by Hokusai, Katsushika. This inspired Canadian Jeff Wall: "A Sudden Gust of Wind".
A flying carp can make the direction of the wind visible. (There's more to it than just the direction...also wind speed.)
What can we use to make the direction of the gravitational force visible?
A "plumb bob" (What's the point of doing this in golf?)
Forest of vectors
direction and magnitude: vector.
Vectors
direction; arrows; magnitude: brightness.
Field lines
direction: line (tangent); magnitude: ??
$\Rightarrow |\myv E|$ is large where lines are close together, and small where lines are far apart.
And so, for an isolated charge lines should radiate equally in all directions.
What's the electron's deviation from its straight path at the end of the field-containing region?
Use...
The problem with determining $\myv E$ from a measurement of the force $\myv F$ on the balloon....
If you had instead measured $\myv F$ on a 'point charge' $q$.... do you need to know the distance to the VdG to figure $\myv E$????
The strange situation of "the" Electrical field if
you draw the field of two point charges. What field is the second charge reacting
to???
Dipole moment:
$$\myv p = q\myv d$$
with $\myv d$ pointing from - to +.