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Lecture 6 E interactions matter Insulator Conductor Ionized Solution Metal Mobile Electrons Initial Drift Velocity: Collision: Δt electron mobility Intermediate:Eventually
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A. E at B due to positive surface charge B. E at B due to all charges Resultant field at B = 0 C. E at B due to all surface charges At B: D. E A due to all charges, due to all surface charges Verify Following Answer: Both lead to: + =
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Example HW problem Example: let Q < 0 For a conducting sphere with charge Q on it... Place positive q 1 at P, get F 1 Place positive q 2 at P get F 2 If shows that then Reasoning: q 1 leads to E 0 q1 p 1
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Example Polarizability Find q, At midpoint of the pipe:
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![If I put no excess charge on the conductor, the net charge on its inner surface will be: a] 0 b] +Q c] -Q d] -2Q e] cannot determine.](/16/4951174/big_thumb.jpg "If I put no excess charge on the conductor, the net charge on its inner surface will be: a] 0 b] +Q c] -Q d] -2Q e] cannot determine.>")
