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
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter25: Gauss’s Law
Section: Chapter Questions
Problem 43PQ: The nonuniform charge density of a solid insulating sphere of radius R is given by = cr2 (r R),...
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![Three very wide insulating plates are placed equally
spaced from each other as shown in the figure. Sheets
have charge densities of o, = + 5 µC/m4, 02 =-5 µC /
m², 03 = + 15 µC / m2. Calculate the electric field vector
in region I.
NOTE: E0=8.85x10-12
C²/ Nm²
a)
Ej=7.24x10 (N/C) and its direction is – x
b)
E;=6.14x10F (N/C) and its direction is – x
Ej=8.47x10 (N/C) and its direction is +x
e)
E,=8.47x10= (N/C) and its direction is – x
Boş bırak](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3cd51eb9-feef-40d5-958e-572867ee7d68%2F1e21041c-e77e-4b0e-8ed4-be91d24ff9c3%2Fhkiffae_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Three very wide insulating plates are placed equally
spaced from each other as shown in the figure. Sheets
have charge densities of o, = + 5 µC/m4, 02 =-5 µC /
m², 03 = + 15 µC / m2. Calculate the electric field vector
in region I.
NOTE: E0=8.85x10-12
C²/ Nm²
a)
Ej=7.24x10 (N/C) and its direction is – x
b)
E;=6.14x10F (N/C) and its direction is – x
Ej=8.47x10 (N/C) and its direction is +x
e)
E,=8.47x10= (N/C) and its direction is – x
Boş bırak
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