hw06 - Physics 21 Spring, 2011 Solution to HW-6 24-28...

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Unformatted text preview: Physics 21 Spring, 2011 Solution to HW-6 24-28 Figure 24-42 shows a thin plastic rod of length L = 13 . 3 cm and a uniform positive charge Q = 55 . 2 fC lying on an x axis. With V = 0 at infinity, find the electric potential at point P 1 on the axis, at distance d = 3 . 60 cm from one end of the rod. To find the electric potential, it will be necessary to integrate over the charge distribution using this formula from your equation sheet: dV = 1 4 dQ | r- r | The necessary steps are to define r and r , define dQ in terms of a variable you can integrate over, and lastly, integrate. The vectors r and r point to your location of interest and source charge dQ respectively. The distance between them is the length | r- r | . Ill choose x as a variable to integrate over that points to the source charge dQ . dQ = dx = Q L dx r =- d i r = x i | r- r | = ( d + x ) Now these quantities can be inserted in to the potential equa- tion: dV = 1 4 Q L dx ( d + x ) Integrate this expression over the length of the rod. V = 1 4 Q L integraldisplay L dx ( d + x ) = 1 4 Q L (ln ( d + L )- ln ( d )) V = 1 4 Q L ln parenleftbigg d + L d parenrightbigg Remember that the units are in volts. 24-34 Two large parallel metal plates are 2 . 2 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero....
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hw06 - Physics 21 Spring, 2011 Solution to HW-6 24-28...

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