Equipotential Lines and field mapping

AI Thread Summary
Equipotential lines are always perpendicular to electric field lines due to the nature of electric forces and potential energy. When moving along an equipotential surface, no work is done, as the potential energy remains constant. The discussion also touches on the relationship between total mechanical energy, potential energy, and kinetic energy in the context of electric fields. Additionally, it raises questions about gravitational fields and the direction of force lines. Understanding these concepts is crucial for accurately mapping electric fields and writing lab reports.
demode
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Equipotential Lines!

I recently did a lab in class that dealt with electric field mapping (very similar to http://physics.nku.edu/GeneralLab/211 Elect Pot. & Field Map.html) and i have to write a lab report now.. I don't understand why the equipotential lines are always perpendicular to the lines of force that were plotted... can someone help?
 
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Thread moved to homework forums. demode, remember to post homework questions (and this lab qualifies as homework) in the correct Homework Help forum.

On your question, what else are the equipotential lines perpendicular to? Also, remember the definition of total mechanical energy TE = PE + KE (sum of potential & kinetic energy). If you are on an equipotential surface and work is done on you, what would happen? What can do work on you?

Quiz question -- what are the equipotential surfaces for a gravitational field (like for us on the surface of the Earth)? Which way do the lines of gravitational force point, and why?
 
Sounds like a homework question, so I'll refrain from fully disclosing the answer.

Consider a moving charge, what direction of movement will result in work being done on the charge? Consider the case of moving perpendicular and parallel to the equipotential lines and then the general case of moving in any direction.

Claude.
 
"Equipotential"- equal potential energy- means that no force has done work to increase of decrease potential. But the work done along a path is the integral of the component of force on that path.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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