Think about what would happen if a positive test charge were placed inside the hollow sphere; it would have no place to escape; remember when drawing electric fields that positive charges need to have field lines drawn outward into a negative source or toward infinity; however, they don't have any place to escape to infinity inside a hollow sphere.
Another way to think about it is that inside a hollow conducting sphere, charges spread themselves out. So, forces pulling or pushing at the center of the sphere cancel out, and the net force is zero. E = F/q, and if F is zero, then E is zero, no matter the charge q.
2007-04-16 04:38:43
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answer #1
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answered by J Z 4
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No charge can exist on the inside of a conductor. According to Gauss theorem ,if u construct a Gaussian sphere anywhere inside or ON the sphere, charge enclosed is 0 . So, field is 0.
2007-04-16 05:09:29
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answer #2
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answered by bua 1
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as the charges reside on the surface of the hollow sphere and not inside,electric field inside it is zero
2007-04-18 22:07:22
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answer #3
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answered by cool dude 1
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you do no longer supply a superb type of element, yet while it incredibly is a charged sphere, then via Gauss' regulation, there can not be an field interior the sector as there is not any internet electric powered cost interior the sector. in spite of if it incredibly is a engaging in sphere, you have an identical argument different than that ifthere is an utilized exterior electric powered field, the exterior costs brought on on the sector are back no longer interior it so there is not any field.
2016-10-22 07:44:56
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answer #4
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answered by ? 4
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This is only true if charge is evenly distributed across the surface of the sphere. The reason it holds true is that if you were to integrate the contribution to the electric field by each arbitrarily small segment of the sphere's surface, you would find that it completely cancels out and the net result is zero.
2007-04-16 04:39:23
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answer #5
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answered by DavidK93 7
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yes According 2 Gauss theorem no charge can B hold inside the conductor
2007-04-16 05:29:55
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answer #6
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answered by Simran 3
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Here is one explanation, hope it is sufficient:
http://orca.phys.uvic.ca/~tatum/elmag/em1.pdf
2007-04-16 04:42:07
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answer #7
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answered by Anonymous
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