39. (a) Let`=0.15 m be the length of the rectangle andw.050 m be its width. Chargeq1is a distance`from pointAand chargeq2is a distancew, so the electric potential atAisVA=14πε0hq1`+q2wi=(8.99×109N·m2/C2)·−5.0×10−6C0.15 m+2.0×10−6C0.050 m¸=6.0×104V.(b) Chargeq1is a distancewfrom pointband chargeq2is a distance`, so the electric potential atBisVB=140hq1w+q2`i=(8.99×109N·m2/C2)·−5.0×10−6C0.050 m+2.0×10−6C0.15 m¸=−7.8×105V.(c) Since the kinetic energy is zero at the beginning and end of the trip, the work done by an externalagent equals the change in the potential energy of the system. The potential energy is the productof the chargeq3and the electric potential. If
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.