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6 Figures

Inline Figure Links

Inline Figure Links

(Will adapt this to Pressbooks, just wanted to get this done and Pressbooks kept rendering the caption tags instead of displaying the text.)

Turning Embedded Activites (Iframes, PHET, etc.) into Figures

Figure_PHET_Guide

(Will adapt this to Pressbooks)

Attaching Captions to Figures

Correct

The caption text will be directly under the figure, and it will match the figure width. It will also be centered.

Four stages of xerography are shown. A positively charged aluminum drum is shown which is grounded. In second stage image is being transferred to it, creating positive image. In third stage, negatively charged toner is attached with the drum and in fourth stage, toner is pulled by the paper which is highly charged.
Figure 9.2: Xerography is a dry copying process based on electrostatics. The major steps in the process are the charging of the photoconducting drum, transfer of an image creating a positive charge duplicate, attraction of toner to the charged parts of the drum, and transfer of toner to the paper. Not shown are heat treatment of the paper and cleansing of the drum for the next copy.

Incorrect

The caption text will be regular paragraph text above or below the image.

The relationship between [latex]V[/latex] and [latex]E[/latex] for parallel conducting plates is [latex]E=V/d[/latex]. (Note that [latex]\Delta \mathit{V}={V}_{\text{AB}}[/latex] in magnitude. For a charge that is moved from plate A at higher potential to plate B at lower potential, a minus sign needs to be included as follows: [latex]–\Delta V={V}_{\text{A}}–{V}_{\text{B}}={V}_{\text{AB}}[/latex]. See the text for details.)

The figure shows two vertically oriented parallel plates A and B separated by a distance d. The plate A is positively charged and B is negatively charged. Electric field lines are parallel between the plates and curved at the ends of the plates. A charge q is moved from A to B. The work done W equals q times V sub A B, and the electric field intensity E equals V sub A B over d and potential difference delta V equals q times V sub A B.

Remediation

First, copy the caption text (then delete it).

Then, click on the image and then the pencil icon, shown below:

Then, paste the caption text into the “Caption” input field:

At the beginning of the caption, add the figure number (the chapter number will be the first number, and the number image it is on the page will be the second). For example, if the chapter number is 12 (shown below) and the image is the first on the page, the figure number is 12.1.

Finally, to make editorial’s job a little easier (and because consistency/readability is important for accessibility too!), center the image:

It should look like this:

The figure shows two vertically oriented parallel plates A and B separated by a distance d. The plate A is positively charged and B is negatively charged. Electric field lines are parallel between the plates and curved at the ends of the plates. A charge q is moved from A to B. The work done W equals q times V sub A B, and the electric field intensity E equals V sub A B over d and potential difference delta V equals q times V sub A B.
Figure 12.1: The relationship between [latex]V[/latex] and [latex]E[/latex] for parallel conducting plates is [latex]E=V/d[/latex]. (Note that [latex]\Delta \mathit{V}={V}_{\text{AB}}[/latex] in magnitude. For a charge that is moved from plate A at higher potential to plate B at lower potential, a minus sign needs to be included as follows: [latex]–\Delta V={V}_{\text{A}}–{V}_{\text{B}}={V}_{\text{AB}}[/latex]. See the text for details.)

License

Icon for the Creative Commons Attribution-NonCommercial 4.0 International License

MSUL's OER Program Internal Processes Copyright © by Chandlee Taylor; Joshua Newman; Julie Taylor; Mary Van Newkirk; and Linda Miles is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted.

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