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

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Physics 21 Spring, 2011 Solution to HW-6 24-28 Figure 24-42 shows a thin plastic rod of length L = 13 . 3cm and a uniform positive charge Q = 55 . 2fC 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 . 60cm 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 πǫ 0 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 | . I’ll choose x as a variable to integrate over that points to the source charge dQ . dQ = λdx = Q L dx r = - d ˆ i r = x ˆ i | r - r | = ( d + x ) Now these quantities can be inserted in to the potential equa- tion: dV = 1 4 πǫ 0 Q L dx ( d + x ) Integrate this expression over the length of the rod. V = 1 4 πǫ 0 Q L integraldisplay L 0 dx ( d + x ) = 1 4 πǫ 0 Q L (ln( d + L ) - ln( d )) V = 1 4 πǫ 0 Q L ln parenleftbigg d + L d parenrightbigg Remember that the units are in volts. 24-34 Two large parallel metal plates are 2 . 2cm 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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