Ranking E field in equipotential surfaces, confused

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SUMMARY

The discussion focuses on ranking electric fields in equipotential surfaces based on their configurations as illustrated in Figure 24-25. The conclusion is that the electric field strengths are ranked as E1 > E2 = E3, where E1 is the strongest and E3 is the weakest. The electric field direction is determined by the potential difference, with E1 and E2 directed upwards and E3 directed downwards. The relevant equation for calculating the electric field is E = -dV/dx, which indicates that the electric field points from regions of higher potential to lower potential.

PREREQUISITES
  • Understanding of electric fields and equipotential surfaces
  • Familiarity with the equation E = -dV/dx
  • Knowledge of potential difference and its relation to electric field direction
  • Basic concepts of charged particles and their behavior in electric fields
NEXT STEPS
  • Study the relationship between electric field strength and equipotential surfaces
  • Learn about the implications of electric field direction in various configurations
  • Explore the concept of work done by electric fields on charged particles
  • Investigate advanced topics in electrostatics, such as Gauss's Law
USEFUL FOR

Students and educators in physics, particularly those studying electromagnetism, as well as anyone seeking to deepen their understanding of electric fields and equipotential surfaces.

mr_coffee
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Hello everyone I'm confused on this topic. I read about it in the book and it made sense though. The question is: Figure 24-25 shows three sets of cross sections of equipotential surfaces; all three cover the same size region of space.
Diagram: http://www.webassign.net/hrw/25_29.gif

(a) Rank the arrangements according to the magnitude of the electric field present in the region, greatest first (use only the symbols > or =, for example 1=2>3).


(b) In which is the electric field directed down the page?


Well in the book it says, E is always perpendicular to the equipotential surfaces and I get that, also i understand that if a charged particle moves from one end to the other but still ends up on the same equipotential surface, the work done is 0. Also if you moved a charged particle to a difference surface, and u moved another charged particle to the same surface, the work done would be equal to each other, because they are both on the same equipotential surface. But for part (a) I'm confused, what equation would I use to calculate the E field, do i assume there is a charged particle moving from each surface to the next until it hits the last one or what? Also for part (b), wouldn't they all be pointing downwards? if i assumed there was a charged particle star4ting at the top of the surfaces, because a + charges E field is always directing away from it, wouldn't they all be down the page?:bugeye:
 
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E = - dV/dX = - \Delta V/ \Delta x for unidirectional field, gives that potential decrease in the direction field hence field is upward in 1 and 2 but downward in 3.

for first
\ E_1 = - (80 -100)/x = 20/x
for second
\ E_2 = - (-120+100)/2x = 20/2x
for third
\ E_3 = - (-30 + 50)/2x = - 20/2x
so 1>2=3
 

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