The potential is the precise sum (calculus: the integral) of all the F `dr = -k q / r^2 `dr contributions, from an infinite distance to distance r, using an integral.

• For University Physics students, note that the integral will have limits infinity to r.
• While we have looked at potential energy changes along a straight line, it turns out that the only things that affects the electrostatic potential and therefore potential energy changes in the vicinity of an isolated point charge q are the initial and final distances from the charge.

The electrostatic potential V(r) at a point is thus V(r) = k q / r.

• Note that if q is positive, then a smaller value of r makes the potential more positive.

• This is as it should be, because if we move a positive test charge closer to a positive charge q, we must do work on it and its potential energy increases. Therefore it electrostatic potential, defined as its potential energy per unit charge, becomes greater.

The potential difference from point A at distance a to point B at distance b is simply the difference `dV in the values of the potential function V at the two points. That is, `dV = V(b) - V(a) = k q / b - k q / a.

• If B is closer to a positive charge q than A, then b < a and k q / b > k q / a, making the difference in potentials positive, as we would expect.

Video Clip #02

In the figure below we determine the potential difference from the point A and a distance of 120 cm from a positive charge point of 8 * 10^-8 C, to point B at a distance of 80 cm from this charge.

• The potential at A is V_A = V(a) = k q / r_A, while the potential at B is V_B = V(b) = k q / r_B.
• These quantities are easily determined, as indicated below.
• For the present configuration we find that V_A = 600 volts while V_B = 900 volts.
• We thus determine that the potential difference from point A to point B is V_B - V_A = 900 V - 600 V = 300 V.
• This 300 V = 300 J / C potential difference is the work per unit charge required to move charge from point A to point B.
• Since the 8 * 10^-8 C charge is positive, this work will be positive. This is consistent with the + 300 V result just obtained.

Video Clip #03

The figure below depicts a system of 3 charges. We wish to find the net force on the 7 `microC charge.

• In order to determine this net force we will determine the vector force F₂₁ of the 5 `microC charge on the 7 `microC charge, and the force F31 of the -12 `microC charge on the 7 `microC charge. We will then add the two forces as vectors to determine the net force on the 7 `microC charge.

We total the x and of the y components of the 2 vectors to obtain the x and the y components of the resultant vector.

• We then determine the magnitude and angle of the resultant force.
• The magnitude of the force has not been determined in the figure below. You should use the Pythagorean Theorem to determine for yourself.  The result will be near .9 N.

Video Clip #04

We can easily determine the energy of this system.

We imagine assembling the system one charge at a time, starting with a system in which no charges are present.

• The first stage of assembly consists of simply placing the charge q₂ at the origin. Since no charges are initially present, we need do no work to place q₂ at this point.
• The second stage consists of bringing in the charge q₁. Since q₂ is now present, there will be force, and hence work, involved in this process.

• We assume that q₁ is brought in from a great, essentially infinite, distance.
• Since q₁ ends up at point A, which is a distance r₁₂ = `sqrt(.65) m from q<sub>2</sub>, the electrostatic potential at A will be V<sub>A</sub> = k q<sub>2</sub> / r<sub>12</sub>.
• Since the electrostatic potential V<sub>A</sub> is the work per unit charge required to bring a charge from infinite distance to position A, we see that the potential energy of charge q<sub>1</sub> at this position must be PE<sub>21</sub> = q<sub>1</sub> V<sub>A</sub> = k q<sub>1</sub> q<sub>2</sub> / r<sub>12</sub>.
• If we evaluate V<sub>A</sub> we see that is approximately 75,000 volts.
• PE<sub>21</sub> is therefore equal to 75,000 volts * 7 `microC = .53 J.

• PE₂₁ is the work required to assemble charges q₂ and q₁ in their indicated positions, and is the potential energy of the system at the end of the second stage of its assembly.

The third stage of the assembly consists of bringing the charge q₃ from a great distance to the indicated point.

• As q₃ is brought into the system, at every point it experiences a net force which is the vector sum of the forces from q₂ and from q₁. It follows that the work required to bring q₃ to the indicated point is equal to the sum of the work that would be required if only charge q₂ was present, and the work the would be required if only charge q₁ was present.
• We therefore determine the potential V_B by adding the potential energies PE₂₃ and PE₁₃.
• We determine PE₂₃ and PE₁₃ as indicated below, with the indicated result.

The total potential energy is therefore as indicated in the figure below.

• We see that the total potential energy of the system is just the sum of the potential energy of every possible pair of charges in the system.

We see that the total energy of the system is negative.

• If the total energy of the system was positive, we would have to do positive work in order to assemble the system. In this case, energy would be released by the system during its disassembly.
• Since the total energy the system is negative, we have to do negative work in order to assemble the system. This means that we could actually get positive work out of the system during assembly, and it would take work to disassemble the system (i.e., to remove the charges to essentially infinite distances from one another).