1917 Hawkins Electrical Guide - Copyright



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English: Suppose that the conditions are as in the figure that is the segment A1 is positive and the segment B1 negative. Now, as A1 moves to the left and B1 to the right, their potentials will rise on account of the work done in separating them against attraction. When A1 comes opposite the segment B2 of the B plate, which is now in contact with the brush Y, it will be at a high positive potential, and will therefore cause a displacement of electricity along the the conductor between Y and Y1 bringing a large negative charge on B2 and sending a positive charge to the segment touching.

As A1 moves on, it passes near the brush Z and is partially discharged into the external circuit. It then passes on until, on touching the brush X it is put in connection with X, and has a new charge, this time negative, driven into it by induction from B2. Positive electricity, then, being carried by the conducting patches from right to left on the upper half of the A plate, and negative from left to right on its lower half.

Hawkins, N. (1917). 
Hawkins electrical guide  
Questions, answers & illustrations
New York: T. Audel & Co.
Page 26
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  • 2007-05-21 13:43 Reddi 481×500×8 (92374 bytes) Hawkins, N. (1917). Hawkins electrical guide Questions, answers & illustrations New York: T. Audel & Co. Page 26

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