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Consider two infinite insulating sheets of charge as shown edge-on in Fig. 1 below. One, with charge per unit area o, is positioned at r=-d, and the other, with charge per unit area 20, is positioned at r +d. Fig. I +20 (a) Find the electrie field vector E for all values of z. Now an electrically neutral, infinite conducting slab of thickness d is inserted in between the charged sheets, as shown below in Fig. 2. +20 Fig. 2

Consider two infinite insulating sheets of charge as shown edge-on in Fig. 1 below. One, with charge per unit area o, is positioned at r=-d, and the other, with charge per unit area 20, is positioned at r +d. Fig. I +20 (a) Find the electrie field vector E for all values of z. Now an electrically neutral, infinite conducting slab of thickness d is inserted in between the charged sheets, as shown below in Fig. 2. +20 Fig. 2

College Physics
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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Consider two infinite insulating sheets of charge as shown edge-on in Fig. 1 below. One, with charge per unit area o, is positioned atr= -d, and the other, with charge per unit area 20, is positioned at r = +d. y Fig. I +20 (a) Find the electric field vector E for all values of z. Now an electrically neutral, infinite conducting slab of thickness d is inserted in between the charged sheets, as shown below in Fig. 2. +20 Fig. 2 *- +d2 -+d (b) Find the location and values of any induced charge distributions in or on the slab. (c) Recompute the electric field vector E for all values of z for this new configuration. (d) Plot the clectric potential V(r) on the x-axis, for all values of r for this configuration. Set the zero of the electric potential at r = 0, i.e. take V(r = 0) = 0.
Transcribed Image Text:Consider two infinite insulating sheets of charge as shown edge-on in Fig. 1 below. One, with charge per unit area o, is positioned atr= -d, and the other, with charge per unit area 20, is positioned at r = +d. y Fig. I +20 (a) Find the electric field vector E for all values of z. Now an electrically neutral, infinite conducting slab of thickness d is inserted in between the charged sheets, as shown below in Fig. 2. +20 Fig. 2 *- +d2 -+d (b) Find the location and values of any induced charge distributions in or on the slab. (c) Recompute the electric field vector E for all values of z for this new configuration. (d) Plot the clectric potential V(r) on the x-axis, for all values of r for this configuration. Set the zero of the electric potential at r = 0, i.e. take V(r = 0) = 0.
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