6. Shown below is a neutral spherical conductor with at total charge Q = +8.00 nC. It has an innerradius of 5.00 cm and an outer radius of 10.0 cm. A charge q = -3.00 nC is placed at the center%3Dof the hole.а.Find the area charge density of the outer surface of the conductor.b.Find the electric field at a point 7.00 cm from the charge.4= πrVsphere = tr³Note: AcircleAsphere = 4ar2= 4ur210.0 cm5.00 cm

(a) Let q denotes the point charge at the center of the hole, Q denotes the net charge on the…

6. Shown below is a neutral spherical conductor with at total charge Q = +8.00 nC. It has an inner radius of 5.00 cm and an outer radius of 10.0 cm. A charge q = -3.00 nC is placed at the center of the hole. а. Find the area charge density of the outer surface of the conductor. b. Find the electric field at a point 7.00 cm from the charge. Note: Acircle = tr² Asphere = 4tr2 Vsphere =tr³ 3 10.0 cm 5.00 cm

6. Shown below is a neutral spherical conductor with at total charge Q = +8.00 nC. It has an inner radius of 5.00 cm and an outer radius of 10.0 cm. A charge q = -3.00 nC is placed at the center of the hole. а. Find the area charge density of the outer surface of the conductor. b. Find the electric field at a point 7.00 cm from the charge. Note: Acircle = tr² Asphere = 4tr2 Vsphere =tr³ 3 10.0 cm 5.00 cm

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter24: Gauss’s Law

Section: Chapter Questions

Problem 24.46P: A thin, square, conducting plate 50.0 cm on a side lies in the xy plane. A total charge of 4.00 108...

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Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

6. Shown below is a neutral spherical conductor with at total charge Q = +8.00 nC. It has an inner
radius of 5.00 cm and an outer radius of 10.0 cm. A charge q = -3.00 nC is placed at the center
%3D
of the hole.
а.
Find the area charge density of the outer surface of the conductor.
b.
Find the electric field at a point 7.00 cm from the charge.
4
= πr
Vsphere = tr³
Note: Acircle
Asphere = 4ar2
= 4ur2
10.0 cm
5.00 cm

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