Potential Difference between two positive point charges

In summary: The electric potential, V, at a point is the potential energy of the charge located at that point. In this case, V is the potential energy of a point charge located at x=a.
  • #1
flannabhra
15
0

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!
 
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  • #2
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!

No, the potential of a charge q located at x=a is kq/|x-a|, notice the absolute value.
 
  • #3
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'.
 
  • #4
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.

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.​
 
  • #5
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.​

I got E = 0 for part (a) but i now feel that it is wrong based on your responses. The variable r represents the distance from the point charge to the Origin.
 
  • #6
flannabhra said:
I got E = 0 for part (a) but i now feel that it is wrong based on your responses.
That is the correct answer for part (a), but do you know why it is?

The variable r represents the distance from the point charge to the Origin.
Distance is a non-negative quantity, as pointed out by Dick. So r can't change the sign of the potential.
 
  • #7
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.

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.
 
  • #8
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.
Yes. That answers the question regarding part (a) very nicely.
 
  • #9
Dick said:
No, the potential of a charge q located at x=a is kq/|x-a|, notice the absolute value.

I don't understand this.. What is x in this situation if r is the distance from the point Q to the origin?
 
  • #10
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'.

Hi!

a) I believe that in this situation, V is scalar.
b) 'r' is the distance from the point Q to the origin
 
  • #11
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?
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.
 
  • #12
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.

So since the Electric Potential is scalar, is V = 2kQ/a correct?

Thank you so much for your help SammyS!
 
  • #13
flannabhra said:
So since the Electric Potential is scalar, is V = 2kQ/a correct?

Thank you so much for your help SammyS!
That's correct.
 
  • #14
My qs is what is the potential difference between two similar charges?
 
  • #15
amirgul2007 said:
My qs is what is the potential difference between two similar charges?
You'll need to be more precise. The potential at a point charge is theoretically infinite (+ or -), point charges being purely theoretical things.
 

1. What is the definition of potential difference?

Potential difference is the difference in electric potential between two points in an electric field. It is also known as voltage and is measured in volts (V).

2. How is potential difference calculated between two positive point charges?

The potential difference between two positive point charges can be calculated by dividing the electric potential energy by the magnitude of charge. It can also be calculated by multiplying the magnitude of the electric field by the distance between the two charges.

3. What is the relationship between potential difference and electric field?

Potential difference and electric field are directly proportional. This means that as the potential difference increases, the electric field strength also increases.

4. Can potential difference be negative?

Yes, potential difference can be negative. This occurs when the direction of the electric field is opposite to the direction of the movement of positive charges.

5. How does potential difference affect the flow of electric current?

Potential difference is the driving force for the flow of electric current. The higher the potential difference, the more energy the charges have and the faster they will move, resulting in a greater flow of current.

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