Electrical Energy and Capacitance41Problem Solutions 16.1(a) The work done is ( )coscosWFsqEsθ=⋅=⋅(), or ( )( )192191.6010C200 N C2.0010m cos06.4010JW−−=××°−(b) The change in the electrical potential energy is 196.4010JePEW−∆=−=− ×(c) The change in the electrical potential is 19-196.4010J4.00 V1.6010CePEVq−∆−×==−×16.2(a) We follow the path from (0,0) to (20 cm,0) to (20 cm,50 cm). The work done on the charge by the field is ( ) ( )() ()121 12 264coscos0.20 m cos00.50 m cos901210C250 V m0.20 m06.010JWW WqqqEθθ=+= ⋅+ ⋅=°+°+=×Thus, 46.010JePEW−
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