Calculating the point where potential V = 0 (due to 2 charges)

Click For Summary
The discussion revolves around the concept of electric potential being zero at two distinct points due to the influence of two charges. It is noted that the equipotential lines around a negative charge can indeed create two points where the potential is zero, one on each side of the charge. The reasoning is supported by the understanding that the potential equation can yield two solutions, particularly in cases involving a quadratic relationship. However, if the charges are equal and opposite, one of the solutions may approach infinity. This highlights the complexity of electric potential calculations in relation to charge placement.
link223
Messages
70
Reaction score
12
Homework Statement
Two point charges, and are placed
5.0 cm apart on the x axis. At what points along the x axis is
(a) the electric field zero and (b) the potential zero? Let
V = 0 at r = infinity.
Relevant Equations
Electric potential
Apparently, there are two solutions where the electric potential is zero which I don't understand, can I get some input on how this is possible?
I have one thing in mind (which I just thought of and might solve it), the equipotentiality i.e. when I draw a circle for V = 0 around the negative charge it will have 2 points where V = 0 one to the right and one to the left because the equipotential lines are continuous. Is that reasoining correct?
1652454851056.png
 
Physics news on Phys.org
I'm not able to read your photo very well. The problem statement does not say what the magnitudes of the 2 charges are.

Also, please have a look at the "LaTeX Guide" link below the Edit window. Posting math equations in LaTeX is *much* more readable that posting dim photos of your hand-written work (and is required by the PF rules). Thanks.
 
link223 said:
Homework Statement:: Two point charges, and are placed
5.0 cm apart on the x axis. At what points along the x-axis is
(a) the electric field zero and (b) the potential zero? Let
V = 0 at r = infinity.
Relevant Equations:: Electric potential

Apparently, there are two solutions where the electric potential is zero which I don't understand, can I get some input on how this is possible?
I have one thing in mind (which I just thought of and might solve it), the equipotentiality i.e. when I draw a circle for V = 0 around the negative charge it will have 2 points where V = 0 one to the right and one to the left because the equipotential lines are continuous. Is that reasoining correct?
View attachment 301419
There must be two solutions in principle because you got a quadratic, and your argument is basically sound, but there will be degenerate cases. E.g. if the charges are equal and opposite then one solution goes to infinity.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
914
Replies
12
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
659
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 30 ·
2
Replies
30
Views
3K