# Potential difference in between positive and negative charge

• Seung Tai Kang
In summary, the problem given involves two point charges, Q1 = +5.00 nC and Q2 = -3.00 nC, separated by a distance of 35.0 cm. The task is to find the electric potential at a point midway between the charges. The equation used for this calculation is V = Ke(q/r), with PE = Vq representing the potential energy, and the standard value of Ke used. The answer, 103J/C, was obtained by substituting the given values into the equation, resulting in Ke(5 nC - 3 nC)/17.5 cm.
Seung Tai Kang

## Homework Statement

Two point charges Q1= +5.00 nC and Q2 = -3.00 nC are separated by 35.0cm. What is the electric potential at a point midway between the charges?

## Homework Equations

V = Ke(q/r), PE = Vq[/B]

## The Attempt at a Solution

This question is from Serway textbook. The answer is 103J/C. I understand the answer was obtained by Ke(5 nC- 3 nC)/ 17.5cm. But I wonder why do we subtract the two? Shouldn't we add them? For example, if +1charge were put at the mid point, the charge would feel both repelled and attracted by positive and negative charges, hence the summative force of the two. But the answer tells that the charge would have the potential energy that is partially canceled out.

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Seung Tai Kang said:
potential difference at the midpoint.
That does not mean anything. You can have the potential at the midpoint (taking potential at infinity to be zero), or the potential difference between two points,
Seung Tai Kang said:
It is actually 205J in my calculation.

How did you do your calculations? To calculate a potential difference, you need two points. Perhaps they are asking for the potential at the midpoint, not the potential difference?

Doc Al said:
How did you do your calculations? To calculate a potential difference, you need two points. Perhaps they are asking for the potential at the midpoint, not the potential difference?

say, for example, +1 charge were near negative charge and to bring the +1 charge to the midpoint the work needed is the sum of the absolute value of 5x10^-9 and -3x10^-9 charges times Ke/r. But to solve for V, we actually take out the absolute value and subtract them. How is it that voltage is not the sum of the abaolute value of the two when the +1charge will feel summative force from the two?

Seung Tai Kang said:
say, for example, +1 charge were near negative charge and to bring the +1 charge to the midpoint the work needed is the sum of the absolute value of 5x10^-9 and -3x10^-9 charges times Ke/r. But to solve for V, we actually take out the absolute value and subtract them. How is it that voltage is not the sum of the abaolute value of the two when the +1charge will feel summative force from the two?
The two points are 5 nano charge and -3 nano charge and 35cm apart. The question asked the potential difference at the midpoint.

Seung Tai Kang said:
near negative charge
How near? At the negative charge the potential would be -∞, and it would take infinite work to move it anywhere.
Seung Tai Kang said:
to bring the +1 charge to the midpoint the work needed is the sum of the absolute value of 5x10^-9 and -3x10^-9 charges times Ke/r.
No, that's wrong. What standard equation are you relying on for that?

Seung Tai Kang said:
The question asked the potential difference at the midpoint.
As Doc Al and I have independently commented, that makes no sense. I would assume they want the potential at the mid point, taking the potential at infinity to be zero. Are you translating from the original?

As requested, please post all of your calculations in detail. From your brief descriptions, it is quite hard to understand what you are doing.

haruspex said:
How near? At the negative charge the potential would be -∞, and it would take infinite work to move it anywhere.

No, that's wrong. What standard equation are you relying on for that?
The equation I used was PE = Vq

It also might be helpful to post the complete problem, literally word for word. (Is it a textbook problem?)

Seung Tai Kang said:
The equation I used was PE = Vq
Ok, but you need to be clear what the variables mean. That gives you the work done in moving a charge q up a potential difference of V. E.g. you could be moving the charge from infinity to a point at potential V. There is nothing there about absolute values.
Now please show how you are using this equation in your calculations.

## Homework Statement

Two point charges Q1= +5.00 nC and Q2 = -3.00 nC are separated by 35.0cm. What is the electric potential at a point midway between the charges?

## Homework Equations

V = Ke(q/r), PE = Vq[/B]

## The Attempt at a Solution

This question is from Serway textbook. The answer is 103J/C. I understand the answer was obtained by Ke(5 nC- 3 nC)/ 17.5cm. But I wonder why do we subtract the two? Shouldn't we add them? For example, if +1charge were put at the mid point, the charge would feel both repelled and attracted by positive and negative charges, hence the summative force of the two. But the answer tells that the charge would have the potential energy that is partially canceled out.

You had posted the above restatement of the problem in a new thread. I moved it to this one. (Don't create multiple threads on the same topic.)

In this revised version, you are asked to find the electric potential, not the potential difference, at the midpoint. That makes sense.

Seung Tai Kang said:

## Homework Statement

Two point charges Q1= +5.00 nC and Q2 = -3.00 nC are separated by 35.0cm. What is the electric potential at a point midway between the charges?

## Homework Equations

V = Ke(q/r), PE = Vq[/B]

## The Attempt at a Solution

This question is from Serway textbook. The answer is 103J/C. I understand the answer was obtained by Ke(5 nC- 3 nC)/ 17.5cm. But I wonder why do we subtract the two? Shouldn't we add them? For example, if +1charge were put at the mid point, the charge would feel both repelled and attracted by positive and negative charges, hence the summative force of the two. But the answer tells that the charge would have the potential energy that is partially canceled out.
If there were a charge of +1 at the midpoint, would it not mean 1C × 103J/C = 103J?

Seung Tai Kang said:
the charge would feel both repelled and attracted by positive and negative charges, hence the summative force of the two
Yes, but that is force, not energy. It relates to the potential gradient, not the potential "height".

## 1. What is potential difference?

Potential difference, also known as voltage, is the difference in electric potential energy between two points in an electric field. It is measured in volts and is a measure of the strength of an electric field.

## 2. What is the difference between positive and negative charges?

Positive charges have more protons than electrons, while negative charges have more electrons than protons. Positive charges are attracted to negative charges and vice versa.

## 3. How does potential difference affect the movement of charges?

Potential difference causes charges to move from areas of high potential energy to areas of low potential energy. This is known as current and is the basis for electricity.

## 4. Can potential difference be measured?

Yes, potential difference can be measured using a voltmeter. The voltmeter measures the difference in potential energy between two points in an electric field.

## 5. What is the relationship between potential difference and electric fields?

Potential difference is directly related to electric fields. The larger the potential difference, the stronger the electric field. This is because a larger potential difference means a greater difference in potential energy between two points, resulting in a stronger force on charges within the field.

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