The discussion revolves around the electric field and electric potential created by two equal positive point charges positioned symmetrically on the x-axis. Participants are examining the calculations related to the electric potential at the origin due to these charges.
There is an ongoing exploration of the concepts involved, with some participants providing clarifications and corrections to earlier statements. The discussion includes attempts to reconcile different interpretations of the variables and the mathematical relationships involved.
Participants are addressing potential misunderstandings regarding the definitions and calculations related to electric potential and electric fields, particularly in relation to the symmetry of the charge configuration.
flannabhra said:Homework Statement
Two equal positive charges Q are fixed on the x-axis, one at +a and the other at -a.
(a) The electric field E at the Origin O
(b) The electric potential V at the origin O
Homework Equations
E=-dV/dr --> V=kQ/r
The Attempt at a Solution
VNet = V0 + V1
I got V0 = -kQ/r (because the vector r is negative)
..and V1 = kQ/r
V = 0
Is this correct?
Thank you for your help!
That's not correct.flannabhra said:Homework Statement
Two equal positive charges Q are fixed on the x-axis, one at +a and the other at -a.
(a) The electric field E at the Origin O
(b) The electric potential V at the origin O
Homework Equations
E=-dV/dr --> V=kQ/r
The Attempt at a Solution
VNet = V0 + V1
I got V0 = -kQ/r (because the vector r is negative)
..and V1 = kQ/r
V = 0
Is this correct?
Thank you for your help!
SammyS said:That's not correct.
First of all, what was your answer for part (a) and how did you get it.For you attempt at part (b):What does the variable, r, represent?
The operation of division is not defined for a vector --- that is to say, division by a vector is not defined.
Electric potential is a scalar quantity.
That is the correct answer for part (a), but do you know why it is?flannabhra said:I got E = 0 for part (a) but i now feel that it is wrong based on your responses.
Distance is a non-negative quantity, as pointed out by Dick. So r can't change the sign of the potential.The variable r represents the distance from the point charge to the Origin.
SammyS said:That is the correct answer for part (a), but do you know why it is?
Distance is a non-negative quantity, as pointed out by Dick. So r can't change the sign of the potential.
Yes. That answers the question regarding part (a) very nicely.flannabhra said:Well a logical explanation sufficed for me(the Electric Fields would be of equal magnitude in opposite directions), but I would like to understand the explanation of it based on formulas.
On the matter of r, I was under the impression that you could take the direction that r is pointing from the reference point and use that as the direction for r. I.e. the equation for E would be:
E = kQ/r3 * r
- in which r is the vector r and the r in the denominator is raised to the third power so as to keep the equation true.
Dick said:No, the potential of a charge q located at x=a is kq/|x-a|, notice the absolute value.
JC Palmer said:Hi Flannabhra!
Let's answer a few questions:
a.) What kind of quantity is the electric potential, V?
b.) Describe to me the significance of 'r'.
For the charge at x = -a, r = ai , where i is the unit vector in the positive x direction.flannabhra said:I don't understand this.. What is x in this situation if r is the distance from the point Q to the origin?
SammyS said:For the charge at x = -a, r = ai , where i is the unit vector in the positive x direction.
For the charge at x = a, r = -ai .
In each case the distance is a, and the vector, r, points away from the charge.
That's correct.flannabhra said:So since the Electric Potential is scalar, is V = 2kQ/a correct?
Thank you so much for your help SammyS!
You'll need to be more precise. The potential at a point charge is theoretically infinite (+ or -), point charges being purely theoretical things.amirgul2007 said:My qs is what is the potential difference between two similar charges?