0
$\begingroup$

I have some questions regarding the potential difference : Question (1) :

The potential difference is defined as the negative of the integral of E.dl with the lower limit being a and the upper limit being b. Does this mean that the physical meaning of the potential difference is the work done against the field ( because of the negative sign) in moving a charge from point a to b ??

Question (2) : Consider we need to find the potential difference between points A and B at radial distances RA and RB from a point postive charge q placed at the origin. This will be equal to the integral of E ( of point charge). dl from RA to RB this will be equal to q/4*pi*efslon ((1/RA)-(1/RB)). If we consider that RB>RA my textbook states that, in this case, "the work done is positive which means that the work done by an external force to move the positive test charge to the original positive charge placed at the origin. This agrees with the fact that two like charges repelling each other". This disagrees with my understanding mentioned above of the potential difference V, which is the definition of V is the work done against the field. Since the work is positive this means that the external source is moving the test positive charge in the direction of the field ( away from the original positive charge ) toward the original positive charge? Please clarify this !

$\endgroup$
0
$\begingroup$

Does this mean that the physical meaning of the potential difference is the work done against the field ( because of the negative sign) in moving a charge from point a to b ?

I'd say that it's the work that would be needed, per unit charge, to move a hypothetical positive charge from point a to point b.

my textbook states that, in this case, "the work done is positive which means that the work done by an external force to move the positive test charge to the original positive charge placed at the origin.

You haven't provided enough context to make it clear whether when they say "the work done" they mean the work the field does on the charge, or the work an external force would need to do on the charge to move it in the field. This could explain the confusion of sign.

In any case, if A is closer to the positive charge generating the field, then the potential at A is higher than the potential at B, and we'd say $V_{ab}$ is positive and $V_{ba}$ is negative.

$\endgroup$
  • $\begingroup$ Thank you for your help. Regarding the first question, what do you mean by "that would be needed" and what do the negative sign in the potential difference equation indicates? . For the second question , it is the work done by an external force. $\endgroup$ – John adams Mar 26 at 19:30
  • $\begingroup$ @Johnadams, my point is that the potential isn't the energy used to move an actual charge. It's a measure of how much energy you would need if you wanted to move a charge that you might someday put there. $\endgroup$ – The Photon Mar 26 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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