2. a) Calculate the potential of inhomogeneously charged finite rod of length 2a with (y) K|y|, where K is a positive real constant. For this goal, you have to adapt the following integral to this problem: = kdq V = r Notice that dq = Ads = X(y')dy' here (y') is a function of the position, and r is given in the picture. A final remark, remember the absolute value function: = y for positive y, and y| = -y for negative y. This will help you during y'-integration over the interval [-a,+a]. b) If you found V(x), then calculate the electric field from the relation E(x)=-dV/dx. c) Plot both functions V(x) and E(x) only for positive x. У dq Ads - y' r = 8

2. a) Calculate the potential of inhomogeneously charged finite rod of length 2a with (y) K|y|, where K is a positive real constant. For this goal, you have to adapt the following integral to this problem: = kdq V = r Notice that dq = Ads = X(y')dy' here (y') is a function of the position, and r is given in the picture. A final remark, remember the absolute value function: = y for positive y, and y| = -y for negative y. This will help you during y'-integration over the interval [-a,+a]. b) If you found V(x), then calculate the electric field from the relation E(x)=-dV/dx. c) Plot both functions V(x) and E(x) only for positive x. У dq Ads - y' r = 8

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A thin rod has uniform charge per length 3w over its length H. The distance between point A and point B is 3H and the distance between point A and point P is 2H. We introduce an integration variables with 5 = 0 chosen to be at point A and the +s direction to be down. The small red segment has length ds and charge dg. We want to find the electric potential at point P. Draw it out--label the all the lengths and the integration variable! A. dV= B. dV C. dV = D. dV= A Which expression below gives the voltage d'V from the small charge dg in the small segment ds? Choose from the choices (A thru F) below: E. dV= F. dV= B с Kdq √√²+4H² Kdg (s² +4H²) Kdq 5 Kdq √(8+3H)² +4H² Kdg ((s+3H)² +4H²) 3.Kdq H ((8+6H)² +97²)³/2 -P +y >+x What are the limits of integration s? [Select] [Select]
4. Having found the voltage difference from knowing the electric field, we can also do the inverse, find the electric field if we know the voltage as a function of position. Since the inverse of integration is differentiation, we have: av Ey ây' The partial derivative OV/Ox means that to take the derivative with respect to x while treating y and z as constant. The electric potential in a region of space is given by Ex = What is the electric field in this region? av Əx' 2 5y V (x, y, z) = V. ((-)² – 57) av əz
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Two large charged plates of charge density ±18µC/m² face each other at a separation of 7 mm. Choose coordinate axes so that both plates are parallel to the xy plane, with the negatively charged plate located at z = 0 and the positively charged plate at z = +7 mm. Define potential so that potential at z = 0 is zero (V(z = 0) = 0). Hint a. Find the electric potential at following values of z: o potential at z = -7 mm: V(z = -7 mm) = o potential at z = +1 mm: V(z = +1 mm): = o potential at z = +7 mm: V(z = +7 mm) = = o potential at z = +9.2 mm: V(z = +9.2 mm) = V. V. V. V. b. An electron is released from rest at the negative plate. With what speed will it strike the positive plate? The electron will strike the positive plate with speed of m/s. (Use "E" notation to enter your answer in scientific notation. For example, to enter 3.14 × 10¹2, enter "3.14E12".)
Please answer parts (a) and (b), showing all work.
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