ECG recording

I understand that the voltage will be positive as measured by the ECG but I was wondering why the voltage reaches a maximum exactly at the midpoint. Is there a physiological reason (like sodium channels closing or something else) that causes the wave to increase, reach a maximum then decrease?

Picture source: Harvard Medical School. Pathophysiology of Heart Disease : A Collaborative Project of Medical Students and Faculty. Edited by Leonard S Lilly, Edition 6 ed., Wolters Kluwer, 2016.


1 Answer 1


The mechanism is depicted in the diagram you posted, most obvious in "A":

ECG (and any voltage measurement) is based on a potential difference

You don't measure the voltage at one spot, you measure the difference between two spots. Those spots are depicted in the diagram with a (-) and (+) symbol, although the (-) is left out of BCD.

In A and D, the charges are arranged such that the (-) and (+) leads 'see' the same charge, so you measure zero voltage difference.

In B and C, there are more negative charges near the (-) lead and more positive charges near the (+) lead: you measure a positive voltage. The maximum in C is because this is the time when the charge measured at (-) and (+) is most different.

Of course the measured ECG is not from one single cell but rather the entire heart, but the concept is the same.

  • This makes a lot of sense, would I be correct in assuming you're using coulombs law between the negative and the positive terminal to essentially get the product of the charges, which is a maximum in the center? Nov 4, 2019 at 22:49
  • @DylanPatel You could use Coulomb's law at each lead, treating the lead as a 'test charge' and calculating the sum force caused by all particles in the system (this is the same as 'measuring the electric field'). The measured voltage is the difference between the electric field at (-) and the electric field at (+).
    – Bryan Krause
    Nov 4, 2019 at 23:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.