Potential Difference between two positive point charges

[flannabhra]

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

V_Net = 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!

 

[Dick]

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

V_Net = 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.

 

[JC Palmer]

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'.

 

[SammyS]

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

V_Net = 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.​

 

[flannabhra]

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.

 

[SammyS]

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.

 

[flannabhra]

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.

 

[SammyS]

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.

 

[flannabhra]

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?

 

[flannabhra]

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

 

[SammyS]

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.

 

[flannabhra]

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!

 

[SammyS]

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

Thank you so much for your help SammyS!

That's correct.

 

[amirgul2007]

My qs is what is the potential difference between two similar charges?

 

[haruspex]

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.