Section 22.2 Charge Diagrams
Subsubsection Key Ideas
Representation 22.2.2. Charge Diagram.
A charge diagram shows a simple picture of an object with small \(+\) and \(-\) labels indicating where charge is present, what type of charge is present, and where the charge density is greatest and smallest.
Example: a wire with dense positive charge on the left, low density charge in the middle, and very dense negative charge on the right.
Subsubsection Activities
Activity 22.2.1. Reading a Charge Diagram.
Shown above is a charge diagram for a disc. Indicate the location with the greatest charge density.
Activity 22.2.2. Diagramming Charge Densities.
Make a charge diagram for each of the following charge densities. What is the purpose of the various constants in these expressions?
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\(\lambda(x) = +\lambda_1 \frac{x}{L}\) between \(x=-L\) and \(x=+L\)
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\(\sigma(s,\theta) = +\sigma_3 \frac{s^2}{R^2}\) between \(s=0\) and \(s=R\text{,}\) where \(s\) is the radial distance from the origin.
Activity 22.2.3. Spreading out Charges.
A wire has four positive point charges located as shown in the figure below.
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What is the total charge on the wire?
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Suppose the wire were placed in a uniform electric field, \(\vec{E} = E_o \hat{y}\text{.}\) What is the net force on the wire?
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Imagine that the total charge on the wire was instead spread uniformly across the wire, from left to right. Does the net force change?
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Now imagine that the total charge was instead spread nonuniformly: much more charge on the right edge of the wire than the left edge. How does your strategy for finding the net force need to change?
Answer.
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The total charge is \(+4q\text{.}\) You can simply add up the charges!
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The electric force is \(\vec{F}^E = q \vec{E}\text{.}\) The total charge on the wire is \(+4q\text{,}\) so the force is \(\vec{F}^E = (+4q)E_o \hat{y}\text{.}\)
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The net force does not change. The same amount of charge is evenly distributed across the wire, just like before (but in smaller chunks of charge), so the force does not change.
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We must integrate the force!\begin{equation*} \vec{F} = q\vec{E} \end{equation*}\begin{equation*} d\vec{F} = \vec{E}dq \end{equation*}Using the linear charge density, we rewrite \(dq\text{:}\)\begin{equation*} d\vec{F} = \vec{E} \lambda(x) dx \end{equation*}Integrating both sides of this expression:\begin{equation*} \vec{F} = \int \vec{E} \lambda(x) dx \end{equation*}Since the electric field does not change in the \(x\) direction, it can be pulled out of the integral:\begin{equation*} \vec{F} = \vec{E} \int \lambda(x) dx \end{equation*}The integral of \(\lambda(x)dx\) gives you the total charge. It turns out, the electric force is the same, because the electric field is uniform and the total charge remained the same!
