# Day 109

Electric potential (V) at any given point in an electric field is defined as the total amount of work required to bring one Coulomb of positive charge from infinity to that point.

V = W/q         or          V = EPE/q

Since the EPE = Work

Electric potential can also be defined as:

V = Σ kq/r

V = electric potential

k = Coulomb’s constant 9 x 109 Nm2/C2

q = source charge

r = distance between source charge and point of electric potential

The Σ sign means “sum of” and is used when the electric potential at a point in space is determined from several source charges.

Electric potential can also be defined as V = Ed

E = electric field strength in N/C or V/m

d = distance from the field source in m

Electric potential is a scalar, that is, it only has magnitude and when calculating the electric potential from several source charges, only straight-line distance from the source charge is needed.

The unit of electric potential is the Volt which is equal to J/C.

Consider a positive source charge and a positive test charge … (repulsive)

To bring the test charge closer to the source charge requires work be done against the field and therefore the electric potential is positive (gain EPE)

The closer the point is to the source charge, the greater the electric potential.

Consider a negative source charge and a positive test charge … (attractive)

To bring the test charge closer to the source charge, work is done by the field and the work is negative (lose EPE)

The closer the point is to the source charge, the smaller the electric potential.

The work required to move a test charge of one Coulomb from one point to another in an electric field is equal to the difference in the electric potential between the two points.   This is referred to as potential difference.

Potential difference is also referred to as voltage.

A battery labeled 9V means that the electric field created by the two charged terminals will do 9 J of work on every Coulomb that is pushed by the battery from one terminal to the other.

Work can therefore be determined by rearranging the electric potential equation to yield:

W = qΔV

In other words, the work done in moving a charge from one potential to a second potential will be equal to the change in potential times the value of the charge being moved.

All charges exist on the outside of a conducting material, in the center there is no electric charge and therefore no electric field.

There is no potential difference between two points on the surface of a charged conductor and therefore no voltage. If a bird lands with both feet on the same high voltage wire, will he get shocked?

## Sample Electric Potential Problems

Example #1: Three charged particles are located at points A, B, and C on a straight line. A is at x = -3.0 m, B is at x = 0 m, and C is at x = 3.0 m. A carries a charge of 18μC, B carries a charge of-12μC, and C carries a charge of 18μC. Suppose that the electric potential far away from the charges is 0 V. Compute the value of the electric potential at a point D located at y = -4m (4 m directly below point B).

Solution #1: Since electric potential is a scalar we only need the distances that each charge is from point D. Setting up a right triangle, we find the common 3-4-5 right triangle. Using the equation V = Σ kq/r, we easily substitute the values of the three charges and distances.

V = 9 x 109Nm2/C2[(18 x 10-6 C/5 m) +(-12 x 10-6 C/4 m) + (18 x 10-6C/5 m)]

V = 3.78 x 104 V

Note: The negative sign of the charge is important and cannot be disregarded in electric potential calculations. Positive charges increase electric potential and negative charges decrease electric potential.

Example #2: Using the same problem setup as in Example #1, now add point E which is located 3 m to the left of A (at x = -6.0 m). How much work is required to move a particle of charge 3.0μC from the point D to the point E?

Solution #2: In order to find the work done in moving the charge we must first know the potential at point D and point E. We already know that the potential at point D is 3.78 x 104 V. We must find the potential at point E using the same method as in Example #1.

VE = Σ kq/r

VE = 9 x 109 Nm2/C2[(18 x 10-6 C/3 m) + (-12 x 10-6 C/ 6 m) + (18 x 10-6C/9 m)

VE = 5.4 x 104 V

The work done is equal to the change in potential times the test charge or…

W = q Δ V

W = (q)(VE – VD)

Remember Δ means (final minus initial)

W = (3.0 x 10-6 C)(54000 V -37800 V)

W = 0.0486 J

Example #3: The point F is located 4 m above B on the y axis. How much work is required to move a particle of charge 3.0μC from the point D to the point F?

Solution #3: The work done in moving the charge is equal to the charge times the change in potential. Since point F is symmetrically at the same point as point D, the electric potential at point D and at point F should be the same. Therefore, the change in potential is zero, and no work is done in moving the charge. Remember, the path taken is independent of the end result. Work is done in moving the charge from Point D to Point B and the opposite sign of work is done in moving the charge from Point B to Point F, but they cancel because of the signs.

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