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Question
Figure 18.67 below represents the electric field in the vicinity of three small charged objects, R, S, and T. The objects have charges −q, +2q, and −q, respectively, and are located on the x-axis at −d, 0, and d. Field vectors of very large magnitude are omitted for clarity.
1. Briefly describe the characteristics of the field diagram that indicate that the sign of the charges of objects R and T is negative and that the sign of the charge of object S is positive.
2. Briefly describe the characteristics of the field diagram that indicate that the magnitudes of the charges of objects R and T are equal and that the magnitude of the charge of object S is about twice that of objects R and T.
For the following parts, an electric field directed to the right is defined to be positive.
1. On the axes below in Figure 18.68, sketch a graph of the electric field E along the x-axis as a function of position x.
2. Write an expression for the electric field E along the x-axis as a function of position x in the region between objects S and T in terms of q, d, and fundamental constants, as appropriate.
3. Your classmate tells you there is a point between S and T where the electric field is zero. Determine whether this statement is true, and explain your reasoning using two of the representations from parts (a), (b), or (c).
1. i) S is positive since field lines point away. T&R are negative since field lines point towards them.
ii) Vecgtors around R & T are the same length beginning at the same distance, so R&T must have the same charge. Vectors of a given length around S start at 6 units away, whereas those of the same length around R&T begin at 4 units away. See video solution for why this means S has twice the charge magnitude of R&T.
2. See video for graph.
3. $E_{net} = kq \left ( \dfrac{2}{x^2} - \dfrac{1}{(x+d)^2} + \dfrac{1}{(d-x)^2} \right )$
4. No, this is not possible since vectors are always pointing to the right, and because the equation for the net electric field in this region will always be positive. See the solution video for an explanation as to why the equation is always positive.
Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 18, Problem 41 (Test Prep for AP® Courses) (12:07) 