Three very wide insulating plates are placed equally spaced from each other as shown in the figure. Sheets have charge densities of 0 m2, 03 = + 15 µC/ m². Calculate the electric field vector in region I. NOTE: E0=8.85x10-12 C² / Nm² = + 5 µC 7 m², 02 = -5 µC / III Ej=7.24x10 (N/C) and its direction is – x Ej=6.14x10 (N/C) and its direction is – x Ej=8.47x105 (N/C) and its direction is +x E1=8.47x105 (N/C) and its direction is – x Boş bırak

Three very wide insulating plates are placed equally spaced from each other as shown in the figure. Sheets have charge densities of 0 m2, 03 = + 15 µC/ m². Calculate the electric field vector in region I. NOTE: E0=8.85x10-12 C² / Nm² = + 5 µC 7 m², 02 = -5 µC / III Ej=7.24x10 (N/C) and its direction is – x Ej=6.14x10 (N/C) and its direction is – x Ej=8.47x105 (N/C) and its direction is +x E1=8.47x105 (N/C) and its direction is – x Boş bırak

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter6: Gauss's Law

Section: Chapter Questions

Problem 49P: Determine if approximate cylindrical symmetry holds for the following situations. State why or why...

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The field due to two parallel plates, the left with a negative surface charge density of −6.0 C/m2 whilethe sheet on the right has a positive surface charge density of +9.0 C/m2is given below. Suppose the plates were spaced 0.0020 m apart. We found that, using Gauss’s law that, independent of this distance, • In I the net field points to the left and has magnitude |Enet| = 1.69 × 1011 N/C.• In region II, between the plates, the field has magnitude 8.47×1011 N/C and points to the left• In III it points to the right and has magnitude |Enet| = 1.69 × 1011 N/C.
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