Question
Consider two conducting plates, placed on adjacent sides of a square, but with a 1-m space between the corner of the square and the plate. These plates are not touching, not centered on each other, but are at right angles. Each plate is 1 m wide. If the plates are held at a fixed potential difference ΔV, sketch the path of both a positively charged object placed between the near ends, and a negatively charged object placed near the open ends.
Final Answer
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Solution video
OpenStax College Physics for AP® Courses, Chapter 19, Problem 30 (Test Prep for AP® Courses)
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Video Transcript
This is College Physics Answers with Shaun Dychko. We have two plates shown in black that have a potential difference between them let's call this one zero, in which case this plate will have a potential V so the high potential plate is positive and the low potential plate is negative then and the blue lines represent the electric field between the plates and they are making an arc from one plate to the next and the spacing between the blue lines increases with distance the electric field is strongest when the plates are close together here but the electric field is weaker closer to the open ends out here and that weakness of the electric field at the open ends is represented by a larger spacing between the electric field lines. Okay! So we are going to show the path of a positive charge located at the near ends and this path is intentionally not following the blue line exactly. So this positive charge will get repelled from the positive plate and so the force on it at this initial position will be straight down and then as it travels, the force on it is gradually going towards the left but this force represents the direction of the acceleration of the charge but once it gets to a position like here say... it already has so much vertically downward velocity that the acceleration is going to deflect it towards the left but it's going to be going further and faster and faster downwards as well and so it's not going to follow the blue line exactly. And the same kind of story happens with this negative charge near the open ends here it's going to follow an arc going from the negative plate up to the positive plate and not quite following the electric field lines and so I guess you could say that the radius of curvature of the path is going to be larger than the radius of curvature of this field line.