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My textbook has the following question

What is the potential difference between two points that are 50cm and 80cm respectively from a point charge of 2uC?

Unfortunately, the textbook has not taught me how to actually do this sort of question, as I only know that $V = ΔU/q$ which is about energy not distance. Using my own inference, I took the potential difference to be the area under the field-distance graph between the two points as this would essentially be equal to the work done on a charge when it is moved between the two points.

By integrating $E = kQ/r^2$ I came to the equation $V = kQ(1/a - 1/b)$ where a and b are two distances. This gives an answer of 13500V.

However, the textbook answer is 44 000V, which is obtained by directly subtracting the value of E at 80cm from the value of E at 50cm ($V = kQ/0.5^2 - kQ/0.8^2)$. I fail to see how the textbook answer is correct - for example, you do not calculate the amount of work done between two points simply by subtracting the force at the points? Isn't this a similar problem?

Am I correct or is my textbook correct, and why?

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    $\begingroup$ By "2uC", do you mean $2μC=2 \times10^{-6}C$? $\endgroup$
    – Adil Mohammed
    Commented Dec 11, 2021 at 9:11
  • $\begingroup$ Yes, I think the textbook is most likely to be wrong. Also note, that potential difference is the work done by the external force against the electric field, which means $F_{ext}=-F_{electrostatic}$ as they are in opposite directions and if you are considering an unit charge $F_{electrostatic}=E$. So the formula would be $$V=-\int_{i}^f Edx$$ $\endgroup$
    – Adil Mohammed
    Commented Dec 11, 2021 at 9:37

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So, we will use the formula

Potential due to charge = Kq/r

  • where K is the constant with value 9 * 10 ^9 kg⋅m3⋅s−2⋅C−2
  • r is disance from the chage to the point where the potential is to be found
  • q is teh value of charge in coulombs

Now we'll find the potential due it charge 2 * 10^-6 Coulombs at the points 50 cm and 80 cm repectively and subtract those two answers, this will give us the potential differnce between the two points

I've shown the deatiled explaination in the image Image Exlanation For the question

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