# Example Find the work done in moving a 5 μ charge from...

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ENERGY AND POTENTIAL
Point Charge in an External Field To move charge Q against the electric field, a force must be applied that counteracts the force on Q that arises from the field: + E Q F appl F appl = - Q E
Differential Work Done on Moving a Point Charge Against an External Field In moving point charge Q from initial position B over a differential distance dL (to final position A) , the work expended is: dW = F appl dL = QE dL = -Q E d L [J] + E F appl + B (initital) A (final) dL d L gives positive result if charge is forced against the electric field
Differential Work Done on Moving a Point Charge Against an External Field The path is along an electric field line (in the opposite direction), and over the differential path length, the field can be assumed constant.
Forcing a Charge Against the Field in an Arbitrary Direction What matters now is the component of force in the direction of motion. Differential work in moving charge Q through distance dL will be: + E F appl = -Q E + B A Force magnitude is F appl cos( d L dW = F appl cos( dL = Q E d L
Example An Electric field is as E = 6y 2 zax +12xyzay +6xy 2 az V/m. An incremental path is represented by Δ L = -3ax + 5ay 2az μ m. Find the work done in moving a 2 μ C charge along this path if the location of the path is at: (a) P A (0,2,5); (b) P B (1,1,1); (c) P C (-2,-0,3)
Total Work Done over an Arbitrary Path The integral expression for work is completely general: Any shape path may be taken, with the component of force evaluated on each differential path segment.
Total Work Done All differential work contributions along the path are summed to give: + E F appl = -Q E + B (initial) A (final) d L
Total Work Done over an Arbitrary Path The integral expression involving the scalar product of the field with a differential path vector is called a line integral or a contour integral.
Line Integral Evaluation We wish to find: where and using these
Example Find the work done in moving a 5 μ charge from the origin to P(2,-1,4) thru the field E = 2xyzax + x 2 zay + x 2 yaz V/m via the path: (a) straight line segments (0,0,0) to (2,0,0) to (2,-1,0) to (2,-1,4); (b) straight line x = -2y, z = 2x; (c) curve x = -2y 3 , z=4y 2 C
Seatwork 1. Calculate the work done in moving a 4 C charge from B(1,0,0) to A(0,2,0) along the path y = 2 2x, z=0 in the field E =: (a) 5a x V/m; (b) 5xa x V/m; (c) 5xa x + 5ya y V/m.
Seatwork