PhysicsLAB Resource Lesson
Electric Potential Energy: Point Charges

In the following diagram, the central charge equals 10 µC.
Surface A (rA = 3 meters) the potential equals:
 
VA = kQ/r = (9 x 109)(10 x 10-6)/3 = 3 x 104 volts
 
 
 
Surface B (rB = 1 meter) the potential equals:
 
VB = kQ/r = (9 x 109)(10 x 10-6)/1 = 9 x 104 volts
 
If a 2 nC charge were to be brought in from infinity and placed on surface A shown above, the amount of work done on the 2 nC charge would equal
 
W = qΔV = 2 x 10-9(ΔV)
2 x 10-9(VA - V∞)
2 x 10-9(3 x 104 - 0)
6 x 10-5 J
 
We say that the 2 nC charge has gained an electric potential energy of
 
EPEA = 6 x 10-5 J. 
 
By definition, the absolute potential at a position infinitely far from a point charge is defined to be zero.

Similarly, the amount of work done on the 2 nC charge to bring it in from infinity and place it on surface B would equal
 
W = qΔV = 2 x 10-9(ΔV)
2 x 10-9(VB - V∞)
2 x 10-9(9 x 104 - 0)
1.8 x 10-4 J
 
We say that the 2 nC charge has gained an electric potential energy of
 
EPEB = 1.8 x 10-4 J. 
 
The difference between the 2 nC's electric potential energy at B and its electric potential energy at A would represent the work required to move the 2 nC change from a position on surface A to a position on surface B.
 
Wdone by external force = q(ΔVabs) = ΔEPE
 
In essence, positive voltage changes mean that an external agent must do work to move a positive charge to a new position in the field, while negative voltage changes mean that the field would be doing all of the work to move a positive charge to its new position in the field.
 
By definition, charges flow naturally from points of high potential to points of low potential. That is, when free to move, a positive charge would instinctively flow from surface B (high potential) to surface A (lower potential). Because of this, work done by electric fields (that is, when the charge moves along a field line in the direction of the field) results in a charge LOSING electric potential energy - that is, the electrostatic force causes a charge to move to positions of lower potential and less electrical potential energy consequently gaining KE.
 
 
 
 
An analogy can be formed between equipotential surfaces and altitudes from the surface of the Earth. At any given altitude, an object with mass has a certain amount of gravitational potential energy, PEg = mgh where h is measured from some arbitrarily set zero level (usually the base of the hill). An external agent must do work against the gravitational field whenever the object's height, altitude, is increased  - this results in the object gaining PEg. The gravitational field does work on the mass whenever the object's height, altitude, is decreased resulting in the object losing PEg.
 
By comparing the aerial view with the side view, you can tell that when the surfaces are closer together on the left, it signifies that the altitude is changing more rapidly, that is, that the slope of the hill is steeper. But regardless of which side of the hill a person climbs, he will do the same amount of work and gain the same amount of PEg = mgh. Remember that this is the definition of a conservative field.
 
If instead, the aerial view where to be considered to be a series of equipotential surfaces, then the electric field is stronger on the left side than on the right side since the same changes in voltage occur in a smaller distance on the left than on the right. However, regardless of which direction a charge is moved from one surface to another, the same amount of work is done, since the charge gains or loses the same amount of electrical potential energy,
 
Workdone by electric field = - q(ΔVabs ) = ΔKE
 
The fact that these changes are path independent signifies that an electric field is also a conservative field -- that is, the only thing that counts is a comparison of the ending position to the initial position, not the path taken between the two points. Remember, whenever an electric field does work on a charged particle, the particle loses electric potential energy and gains kinetic energy. This is analogous to gravitational fields. When gravity does work on a mass, it loses potential energy and gains kinetic energy.
 
If a system contains more than one charge, then the EPE of the system is the sum or the EPE of each pair of charge.
EPEsys = k Σqiqj/rij
 
In the following collection of charges, if each charge lies at the corner of a square of side s, then what is the EPEsys?


Note that:

r12 = r23 = r34 = r14 = s
r13 = r24 = s 
There are 6 "pairs" in this diagram:
q1q2 q1q4 q2q4
q1q3 q2q3 q3q4

The total electric potential energy of this system would equal

= k Σqiqj/rij
= k (q1q2/r12 + q1q3/r13 + q1q4/r14 + q2q3/r23 + q2q4/r24 + q3q4/r34)
= k (-q2/s + -q2/s + -q2/s + q2/s + q2/s + q2/s + q2/s)
= k ( - q2/s -q2/s + -q2/s q2/s q2/s q2/s + q2/s)
= k (q2/s)
 
Refer to the following information for the next nine questions.

Use the following electric field map to discern if you understand the concepts discussed so far regarding electric fields and electric potentials.
 
 Where is the electric field strongest? L, M, N, R, S, T, U  (Support your choice)

 Where is the electric field weakest? L, M, N, R, S, T, U (Support your choice)

 What is the direction of the electric field at R?

 How much work would it take an external agent to move a charge from R to N?

 What does the negative sign mean in the previous answer?

 Does it take more work to move a 2 µC charge from R to L and then to T compared to going directly to T?

 How much EPE did the 2 µC charge have while it was at rest at position R?

 What is the 2 µC charge’s EPE at point T?

 How much work was required to move the 2 µC charge from R to T?





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