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To make my question clear I've added a picture above which I drew myself which shows 2 scenarios where the only difference is the sign of the charge. Essentially we have an electric field produced by a positive electric plate and it's effects of repulsion/attraction on a positive/negative charge.

Now, to calculate the voltage I simply use the Voltage=Delta Energy/Coulomb formula. I use the Lower potential energy point as the reference point.In the positive charge scenario it gives me +12V (This makes sense as it's at a higher potential energy point relative to the reference point). Whereas the negative charge scenario it gives me -12V (which doesn't make sense since it's at a higher potential energy point RELATIVE to the reference point so it should also be positive).

Do we just take the absolute value of the charge when dealing with Voltage? I hope I explained it well enough. I'd really appreciate some help. (The numbers are arbitrary by the way)

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The potential difference between two points is equal to the work done by an external force in moving unit positive charge between those two points.

So for your top diagram with the positive charge starting on the left, potential $V$, and moving to the right, potential $0$, the work done by an external force on a charge of $+1\,\rm C$ is $-12\rm \,J$.

So the change in potential $0-V=-12 \Rightarrow V=+12$.

In your second diagram moving the negative charge from right to left requires the external force to do $-12\,\rm J$ of work which would be $+12\,\rm J$ of work if the charge were positive.

So the change in potential $V-0=+12 \Rightarrow V=+12$ as before.

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  • $\begingroup$ This makes sense. Thank you very much $\endgroup$
    – ayazasker
    Mar 19, 2020 at 0:10
  • $\begingroup$ Isn't there a drop in potential in both cases? $\endgroup$ Mar 19, 2020 at 2:09
  • $\begingroup$ @sammygerbil Indeed there is a drop as reflected in the term $0-V$ which is negative if $V$ is a positive quantity. $\endgroup$
    – Farcher
    Mar 19, 2020 at 8:32

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