
#1
Feb1809, 08:13 PM


#2
Feb1809, 09:32 PM

P: 27

Anyone?




#3
Feb1809, 09:36 PM

PF Gold
P: 363

Here's some relevant theory. Taking your reference position to be V = 0 at infinity, the electric potential, V, is given by
[tex] V = \frac{1}{4 \pi \epsilon_o} \frac{q}{r} [/tex] This is related to the electric field by [tex] E =  \nabla V [/tex] which for a point charge can be rewritten as [tex] E =  \frac{1}{4 \pi \epsilon_o} \frac{\partial}{\partial r} \frac{q}{r} [/tex] calculating out the derivative, you get [tex] E = \frac{1}{4 \pi \epsilon_o} \frac{q}{r^2} [/tex] The lines drawn on your diagram are lines of constant electric potential. They are analogous to lines of constant altitude on a topographic map. The closer together the equipotential lines are, the larger the gradient, so the stronger the electric field is. The same thing is true on a topographic map steep hills are indicated by contour lines close together. Hope this helps. 



#4
Feb1809, 09:49 PM

P: 27

Equipotential Contour Theory
Thanks.
I still don't know how each would apply to the questions though. A) The electric field at g is zero. > I can't tell whether this is true or not, I'm starting to have my doubts, but I can't explain why. B) Charge Q2 is the largest negative charge. > I was told the diameter of the charge means it has more strength, therefore Q1 is the largest negative charge, so this is definitely false... right? C) Q1 is a negative charge. > True. Unless I've been reading this wrong. D) The electric field at d is stronger than at b. > I thought because they are on the same contour lines, the fields are the same, but again, like A, I don't get the reasoning, but judging by the fact that closer equipotential lines are, then stronger gradients, this must be true. E) Charge Q3 is the largest positive charge. > It's the only positive charge. True. F) The force on an electron at g points to the top of the page. > False, I'm pretty sure on this one now from my reading. 



#5
Feb1809, 09:59 PM

P: 27

Is the electric field zero if the equipotential contour line is 0 volts?
How does the diameter of the circle affect the strength of the charge? Are electric fields on the same contour line equal or judging by the fact that closer equipotential lines are, then stronger gradients, then can they be different? What about the force on an electrons and why they would point up/down? 



#6
Feb1809, 11:04 PM

P: 27

Wait... are the correct answers C, D, and E?
C is true because q1 is a negative charge. D is true because the contour lines are closer together, and D is closer to the positive charge meaning the field must be stronger because... the charge is stronger. E is true because q3 is the only positive charge. A is false because there are positive charges and negative charge and so it naturally cannot be 0. B is false because q1 has a large radius and therefore, looking at the formula rearranged, that would make q larger. F is false because the electron should point down in alignment with its charge. Is this right?? unfortunately Im not given the correct answers and I cant find anything that explains this clearly to me, usually I learn through examples, but I can't find any. Can anyone confirm or disprove my reasoning? 



#7
Feb1909, 08:48 AM

PF Gold
P: 363





#8
Feb1909, 10:24 AM

P: 27

Thanks so much!!!
I understand it perfectly now!!! 


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