# What if there is acceleration for the unit positive charge while measuring the electrostatic potential?

Adding the general definition of electrostatic potential:

" Electrostatic potential (V ) at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point "

What if there is acceleration for unit +ve charge?

• It is unclear what exactly you are asking. Are you asking what would happen if you moved the charge to the same place only faster ? – HsMjstyMstdn Mar 22 '17 at 11:54
• Why in the definition it is mentioned as (Without acceleration). If it is not clear I can redefine the question – vt673 Mar 22 '17 at 11:56
• The acceleration has nothing to do with it, because acceleration is a function of time not energy. You can move the charge really really fast but if it's through the same distance, it doesn't mean you've done more work, it just means you have a more powerful engine. – HsMjstyMstdn Mar 22 '17 at 12:02

## 2 Answers

If there is acceleration of the test charge then the charge would have kinetic energy along with the potential energy.

Potential is defined as the work done by the external force per unit charge against the electric field of the reference charge (the source of electric field). So if the charge is accelerating then it means that the external force is not equal to the force due to the electric field. Thus, we won't get the correct potential this way.

While calculating the potential due a charge, we only consider the change in potential energy of the test charge when bringing the charge from infinity towards the reference charge.

We only want to consider the potential energy per unit charge to calculate the potential, so that if we wish the find the potential energy of an another charge, we just have to multiply the charge of that particle and the potential of the point where we want to find it's potential energy.

• Is that why (in general) while calculating potential energy of anything we do it in infinitesimal steps? – Kunal Pawar Mar 22 '17 at 15:39
• We do that because when the take the test charge closer to the charge generating the electric field, the force due to this electric field changes. So we have to change our external force accordingly for each step. Each step is so small that the magnitude of the external force can be considered equal to the electric field's force. Thus, for each infinitesimal step the external force is equal to the electric force. If we integrate the work done in each infinitesimal step (each step has different external force), we will get the total work done. That's how integral calculus works. – Mitchell Mar 22 '17 at 15:43
• @BhavyaSharma: If there is no acceleration for the charge, then there is no electromagnetic waves, that means there is no electrostatic field and force!. That means no potential energy! I know somewhere I am missing the logic. Please do advice. – vt673 Mar 23 '17 at 11:43
• Static charges produce electric field around them. Charges moving with velocity produce electric fields along with magnetic fields and accelerating charges release EM waves. There is no need for charges to produce EM waves to form Electric field around them. The whole branch of Electrostatics deals with stationary charges. And these static charges are very well capable of producing electric fields around them. Interaction between electric fields of two charges results in electrostatic forces. – Mitchell Mar 23 '17 at 11:53

We consider a test charge $q$, the magnitude of which is infinitesimely small so that it wouldn't affect the electric field due to source charge. Hence we can calculate potential due to large charge (suppose $x$) easily just by multiplying $x$ with $W/q$ (which is the actual potential).

Now If acceleartion comes into play then accelerated charge particle emits electromagnetic radiation or wave; a wave is always an energy carrier which carry vibrational energy, this energy in turn do work by applying force on source charge causing it to displace to another position thereby changing the configuration of original electric field due to source charge. And at this point if we analyse the accelerated charge then it can be well understood that it violates the assumption that the test charge doesn't affect the electric field due to source charge and we can't further proceed to derive the expression of electric potential in the way which is done above. So it is better to do work on test charge by applying external force but without accelerating it.