Charged Wire in Uniform Field
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Exercise:
A charged wire is placed in a uniform electric field. center includegraphicswidthcm#image_path:chargin-uniform-field-# center abcliste abc Derive a formal expression for the force acting on the wire. abc Calculate the magnitude of the force on a wire with charge QO in a field with strength EO. What is the direction of the force? abcliste
Solution:
abcliste abc Since the electric field is uniform the distribution of the charge along the wire does not matter. This means we can treat the charge like a po charge. Hence the force on the wire is F_E FformF For a more rigorous proof we split up the wire o N short segments of length textdell. If the segments are short enough we can consider them to be po charges. The force on segment i is therefore F_i q_i E where q_i is the charge of segment i i dots N. The total force on the wire is then F_E _i^N F_i _i^N q_i E labeleq: E _i^N q_i E Q where Q _i^N q_i is the total charge of the wire. vspacemm Remark: For infinitely small segments textdell to refeq: becomes F_E _^ell E fracQell textdell E _^ell lambdatextdell E lambda _^ell textdell E lambda ell E Q The result is correct for an arbitrary charge distribution in the wire i.e. it does not have to be uniform. abc The magnitude of the force is F_E FF absQtimesE F approx resultFP- Since the charge is negative the force pos in the opposite direction of the field lines. abcliste
A charged wire is placed in a uniform electric field. center includegraphicswidthcm#image_path:chargin-uniform-field-# center abcliste abc Derive a formal expression for the force acting on the wire. abc Calculate the magnitude of the force on a wire with charge QO in a field with strength EO. What is the direction of the force? abcliste
Solution:
abcliste abc Since the electric field is uniform the distribution of the charge along the wire does not matter. This means we can treat the charge like a po charge. Hence the force on the wire is F_E FformF For a more rigorous proof we split up the wire o N short segments of length textdell. If the segments are short enough we can consider them to be po charges. The force on segment i is therefore F_i q_i E where q_i is the charge of segment i i dots N. The total force on the wire is then F_E _i^N F_i _i^N q_i E labeleq: E _i^N q_i E Q where Q _i^N q_i is the total charge of the wire. vspacemm Remark: For infinitely small segments textdell to refeq: becomes F_E _^ell E fracQell textdell E _^ell lambdatextdell E lambda _^ell textdell E lambda ell E Q The result is correct for an arbitrary charge distribution in the wire i.e. it does not have to be uniform. abc The magnitude of the force is F_E FF absQtimesE F approx resultFP- Since the charge is negative the force pos in the opposite direction of the field lines. abcliste