In physics courses, Coulomb’s law is often used as a type of algebraic recipe to solve physics word problems. Three such examples are shown here.

 Example ASuppose that two point charges, each with a charge of +1.00 Coulomb are separated by a distance of 1.00 meter. Determine the magnitude of the electrical force of repulsion between them.

This is not the most difficult mathematical problem that could be selected. It certainly was not chosen for its mathematical rigor. The problem-solving strategy utilized here may seem unnecessary given the simplicity of the given values. Nonetheless, the strategy will be used to illustrate its usefulness to any Coulomb’s law problem.

The first step of the strategy is the identification and listing of known information in variable form. Here we know the charges of the two objects (Q1 and Q2) and the separation distance between them (d). The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the force. So Felect is the unknown quantity. The results of the first two steps are shown in the table below.

 Given:Q1 = 1.00 CQ2 = 1.00 Cd = 1.00 m Find:Felect = ???

The next and final step of the strategy involves substituting known values into the Coulomb’s law equation and using proper algebraic steps to solve for the unknown information. This step is shown below.Felect = k • Q1 • Q/ d2Felect = (9.0 x 109 N•m2/C2) • (1.00 C) • (1.00 C) / (1.00 m)2

Felect = 9.0 x 109 N

The force of repulsion of two +1.00 Coulomb charges held 1.00 meter apart is 9 billion Newton. This is an incredibly large force that compares in magnitude to the weight of more than 2000 jetliners.

This problem was chosen primarily for its conceptual message. Objects simply do not acquire charges on the order of 1.00 Coulomb. In fact, more likely Q values are on the order of 10-9 or possibly 10-6 Coulombs. For this reason, a Greek prefix is often used in front of the Coulomb as a unit of charge. Charge is often expressed in units of microCoulomb (µC) and nanoCoulomb (nC). If a problem states the charge in these units, it is advisable to first convert to Coulombs prior to substitution into the Coulomb’s law equation. The following unit equivalencies will assist in such conversions. 1 Coulomb = 106 microCoulomb 1 Coulomb = 109 nanoCoulomb

The problem-solving strategy used in Example A included three steps:

1. Identify and list known information in variable form.
2. List the unknown (or desired) information in variable form.
3. Substitute known values into the Coulomb’s law equation and using proper algebraic steps to solve for the unknown information. (In some cases and for some students, it might be easier to first do the algebra using the variables and then perform the substitution as the last step.)

This same problem-solving strategy is demonstrated in Example B below.

 Example BTwo balloons are charged with an identical quantity and type of charge: -6.25 nC. They are held apart at a separation distance of 61.7 cm. Determine the magnitude of the electrical force of repulsion between them.

The problem states the value of Q1 and Q2. Since these values are expressed in units of nanoCoulombs (nC), the conversion to Coulombs must be made. The problem also states the separation distance (d). Since distance is given in units of centimeters (cm), the conversion to meters must also be made. These conversions are required since the units of charge and distance in the Coulomb’s constant are Coulombs and meters. The unknown quantity is the electrical force (F). The results of the first two steps are shown in the table below.

 Given:Q1 = -6.25 nC = -6.25 x 10-9 CQ2 = -6.25 nC = -6.25 x 10-9 Cd = 61.7 cm = 0.617 m Find:Felect = ???

The final step of the strategy involves substituting known values into the Coulomb’s law equation and using proper algebraic steps to solve for the unknown information. This substitution and algebra is shown below.Felect = k • Q1 • Q/ d2Felect = (9.0 x 109 N•m2/C2) • (6.25 x 10-9 C) • (6.25 x 10-9 C) / (0.617 m)2

Felect = 9.23 x 10-7 N

Note that the «-» sign was dropped from the Q1 and Q2 values prior to substitution into the Coulomb’s law equation. As mentioned above, the use of «+» and «-» signs in the equation would result in a positive force value if Q1 and Q2 are like charged and a negative force value if Q1 and Q2 are oppositely charged. The resulting «+» and «-» signs on F signifies whether the force is attractive (a «-» F value) or repulsive (a «+» F value).

 Example CTwo balloons with charges of +3.37 µC and -8.21 µC attract each other with a force of 0.0626 Newton. Determine the separation distance between the two balloons.

The problem states the value of Q1 and Q2. Since these values are in units of microCoulombs (µC), the conversion to Coulombs will be made. The problem also states the electrical force (F). The unknown quantity is the separation distance (d). The results of the first two steps are shown in the table below.

 Given:Q1 = +3.37 µC = +3.37 x 10-6 CQ2 = -8.21 µC = -8.21 x 10-6 CFelect = -0.0626 N (use a — force value since it is attractive) Find:d = ???

As mentioned above, the use of the «+» and «-» signs is optional. However, if they are used, then they have to be used consistently for the Q values and the F values. Their use in the equation is illustrated in this problem.

The final step of the strategy involves substituting known values into the Coulomb’s law equation and using proper algebraic steps to solve for the unknown information. In this case, the algebra is done first and the substitution is performed last. This algebra and substitution is shown below.Felect = k • Q1 • Q/ d2d2 • Felect = k • Q1 • Q2

d2 = k • Q1 • Q/ Felect

d = SQRT(k • Q1 • Q2) / Felect

d = SQRT [(9.0 x 109 N•m2/C2) • (-8.21 x 10-6 C) • (+3.37 x 10-6 C) / (-0.0626 N)]

d = Sqrt [ +3.98 m]

d = +1.99 m

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#### Zhakupov Nursultan

Physics teacher in Nazarbayev Intellectual school, Pavlodar.