Force between Uniformly Charged Parallel Wires
Question
Solution
Short
Video
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Two wires with charges pm Q are placed parallel to each other. We ase that the distance d between the wires is small compared to their length. center includegraphicswidthcm#image_path:forcbetween-parallel-wires-# center abcliste abc Derive a formal expression for the force between the wires. abc Calculate the force between the wires for a charge pmQO a length LO and a distance dO. Compare the result to the result predicted by Coulomb's law for two po charges at the same distance. abcliste
Solution:
abcliste abc The field of the positive wire at the position of the negative wire is given by E_+ fracpivarepsilon_fraclambda_+d fracpivarepsilon_fracQell d Since the field is constant along the negative wire both in magnitude and direction the force experienced by the negative wire is F_- abs-Q E_+ Q fracpivarepsilon_fracQell d fracpivarepsilon_fracQ^ell d The direction of the force is in the opposite direction of the field lines going from the positive to the negative wire i.e. towards the positive wire. vspacemm The force acting on the positive wire is equal in magnitude but opposite in direction which is in agreement with Newton's third law actioreactio. abc The force is F FF fracpitimesncepsfracleftQright^Ltimesd F approx resultFP- Coulomb's law for two po charges at the same distance yields F_C FCF fracpitimesncepsfracleftQright^leftdright^ FC approx resultFCP- which is much greater than the force between the two wires. This can be explained with the fact that the partial forces between charges on the positive and charges on the negative wire are much weaker on average larger distance and have a component parallel to the wire that is cancelled because of the symmetry of the setup. abcliste
Two wires with charges pm Q are placed parallel to each other. We ase that the distance d between the wires is small compared to their length. center includegraphicswidthcm#image_path:forcbetween-parallel-wires-# center abcliste abc Derive a formal expression for the force between the wires. abc Calculate the force between the wires for a charge pmQO a length LO and a distance dO. Compare the result to the result predicted by Coulomb's law for two po charges at the same distance. abcliste
Solution:
abcliste abc The field of the positive wire at the position of the negative wire is given by E_+ fracpivarepsilon_fraclambda_+d fracpivarepsilon_fracQell d Since the field is constant along the negative wire both in magnitude and direction the force experienced by the negative wire is F_- abs-Q E_+ Q fracpivarepsilon_fracQell d fracpivarepsilon_fracQ^ell d The direction of the force is in the opposite direction of the field lines going from the positive to the negative wire i.e. towards the positive wire. vspacemm The force acting on the positive wire is equal in magnitude but opposite in direction which is in agreement with Newton's third law actioreactio. abc The force is F FF fracpitimesncepsfracleftQright^Ltimesd F approx resultFP- Coulomb's law for two po charges at the same distance yields F_C FCF fracpitimesncepsfracleftQright^leftdright^ FC approx resultFCP- which is much greater than the force between the two wires. This can be explained with the fact that the partial forces between charges on the positive and charges on the negative wire are much weaker on average larger distance and have a component parallel to the wire that is cancelled because of the symmetry of the setup. abcliste