Calculating PD between +q and -q Charges at Points A & B

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In summary, the conversation discusses the calculation of potential difference between two points, A and B, with charges +q and -q separated by a distance R. The person attempts to solve the problem by integrating the electric field and directly finding the potential at the two points, but both methods yield a result of 2kq/0, which is nonsensical. The value of K is given as 1/4pi e0 and there is confusion over the limits of integration and the necessity of integrating when only two charges are present. The person also shares a working image but cannot find their data cable.
  • #1
Highwaydude
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Homework Statement



Charges +q and -q at points A and B respectively separated by distance R. Required to calculate the potential difference between these 2 points.

Homework Equations



I want to do in two ways.. i.e once by integrating the electric field and the other by directly finding the potential at the two points and then subtract them to have the difference

The Attempt at a Solution


Both of my trials have yielded a 2kq/0 which is a bit absurd...
I do like this>
Vb -Va= - ((integration sign)kq/r^2 dr-(int.sign)kq/(R-r)^2 dr)
between the limits 0 to R.. It yields 2kq/0(wth!)-2kq/R...
So yeah.. I need help. Oh and K = 1/4pi e0
 
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  • #2
please check if you are putting the limit right
 
  • #3
also i don't knw why r u integrating when only two charges are present
 
  • #4
i am calculating the net electring field... i guess the two kq/0's cancel out one being negative, but its just not coming. i mean bot hare coming out to be positive...
 
  • #5
Heres the working... sorry, i can't find my mobile phone data cable.. had to do it in paint... http://hostmypicture.com/images/untitlete.png
 
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1. How do you calculate the PD between +q and -q charges at points A & B?

To calculate the potential difference (PD) between two charges, you can use the formula PD = (k*q)/r, where k is the Coulomb's constant (9x10^9 Nm^2/C^2), q is the magnitude of the charges, and r is the distance between the two points. In this case, you would use the same formula for both point A and point B, and then subtract the two values to find the PD between them.

2. What is the unit of measurement for PD?

The unit of measurement for PD is volts (V). This is because PD is a measure of the potential energy per unit charge at a specific point in an electric field.

3. Can the PD between two charges ever be negative?

No, the PD between two charges can never be negative. This is because PD is a measure of the difference in potential energy between two points, and potential energy cannot be negative. However, the sign of the PD can indicate the direction of the electric field between the two charges.

4. How does the distance between the charges affect the PD?

The PD between two charges is inversely proportional to the distance between them. This means that as the distance between the charges increases, the PD decreases. This relationship is described by the inverse square law, which states that the PD is proportional to 1/r^2, where r is the distance between the two charges.

5. Can PD be calculated for more than two charges?

Yes, PD can be calculated for any number of charges. However, the formula becomes more complex as the number of charges increases. For three or more charges, you would need to use the superposition principle, which involves summing up the individual PDs between each pair of charges to find the total PD.

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