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10 April 2016

Mail time: What's the difference between (qV) and (q*delta-V) in electrostatics?

A reader discovered these two quiz items among the materials I gave him at a summer institute:

The electric potential at point A is -30 V; the electric potential at position B is 0 V.  

1. What is the electric potential energy experienced by an object with charge +1 C when it is placed at point A?

2. What is the electric potential energy experienced by an object with charge +1 C when it is placed at point B?

(Read carefully.  I answer #1 and #2 at the bottom of the post.  But a reader asked something different.)

The question from the reader asked:

"I know the relevant equation is PE = W = q(deltaV).  My issue is with the delta...It seems the answer to #1 is -30 V since deltaV is (-30V). Does this mean for #2 the answer is choice A since deltaV is (+30V) or zero? Are we even using deltaV or just the electric potential for that point, in which case it is zero for #2?"

The delta in the equation referenced is part of the work-energy theorem -- work done by an external force is equal to the change in kinetic plus change in potential energy.  Here, the potential energy experienced by a charge q is qV, where V is the electric potential at a position.  

The equation referenced -- W = PE = q(deltaV) -- answers not this, but a different question! 

The question that I think you mean to answer is, "How much work is necessary to move the object from A to B (with no change in kinetic energy)?"

From simplifying the work-energy theorem, the work necessary to move the charge is qdeltaV = (+1 C)(+30 V) = +30 J.  The value of the electric potential itself is irrelevant... the term that appears in the work-energy theorem is the CHANGE in potential energy.  Work done by an external force CHANGES the object's potential energy.  And to change potential energy, the object has to move from one position to another where the electric potential has changed.

The charged object moved from point A to point B.  Call the electric potential zero at point B-- fine.  There's no electric potential energy at B.  That's actually not relevant to the problem.

What's relevant is how much greater or less the electric potential at B is.  A positively charged particle moving from V = 0 to v = 30 V gained electric potential energy, exactly the same amount as if it had moved from V =100 V to V = 130 V or from V = -30 V to V = 0 V.

And that's why there's a delta in the equation you referenced.

But now to the original quiz question, from the top of the post:

What's the potential energy of the +1 C charge at point A?  When using PE = qV, the negative signs are important.  So PE = (+1 C)(-30 V) = -30 J.  It does have potential energy, even though it's never moved.  But that potential energy is kinda meaningless unless the charge does move.  Kinda like I have potential energy relative to the ground when I stand on top of the Sears tower, but that's rather meaningless unless I jump off.  :-)


The charge has no potential energy at B because PE = qV = (+1 C)(0 V) = 0 J.

1 comment:

  1. The delta in the equation comes from the Electrostatic Field being a conservative vector field. The consequence of it being a conservative vector field is that the line integral of the Electric field is equivalent to evaluating the difference of the scalar field at the end points.

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