The interaction between charged objects is a not-contact force that acts over some distance of separation. Accuse, charge and distance. Every electrical interaction involves a force that highlights the importance of these three variables. Whether it is a plastic golf tube alluring newspaper bits, 2 like-charged balloons repelling or a charged Styrofoam plate interacting with electrons in a slice of aluminum, there is always two charges and a altitude between them as the 3 disquisitional variables that influence the strength of the interaction. In this section of Lesson 3, we will explore the importance of these 3 variables. Force every bit a Vector Quantity
The electrical force, similar all forces, is typically expressed using the unit of measurement Newton. Existence a strength, the force of the electrical interaction is a vector quantity that has both magnitude and direction. The direction of the electrical force is dependent upon whether the charged objects are charged with like charge or opposite charge and upon their spatial orientation. Past knowing the type of accuse on the two objects, the direction of the force on either one of them tin exist determined with a little reasoning. In the diagram below, objects A and B accept like charge causing them to repel each other. Thus, the force on object A is directed leftward (away from B) and the strength on object B is directed rightward (away from A). On the other hand, objects C and D have reverse charge causing them to attract each other. Thus, the strength on object C is directed rightward (toward object D) and the force on object D is directed leftward (toward object C). When it comes to the electrical force vector, peradventure the all-time way to determine the direction of it is to employ the fundamental rules of charge interaction (opposites concenter and likes repel) using a little reasoning.
Electric force also has a magnitude or force. Like well-nigh types of forces, at that place are a variety of factors that influence the magnitude of the electrical force. Two similar-charged balloons will repel each other and the strength of their repulsive force can be altered past changing three variables. Get-go, the quantity of charge on i of the balloons will bear on the strength of the repulsive force. The more charged a balloon is, the greater the repulsive force. Second, the quantity of charge on the second airship will affect the strength of the repulsive force. Gently rub two balloons with animal fur and they repel a little. Rub the two balloons vigorously to impart more accuse to both of them, and they repel a lot. Finally, the altitude betwixt the two balloons volition have a significant and noticeable effect upon the repulsive forcefulness. The electrical force is strongest when the balloons are closest together. Decreasing the separation distance increases the force. The magnitude of the force and the distance between the two balloons is said to be inversely related.
Coulomb'due south Law Equation
The quantitative expression for the result of these three variables on electrical force is known as Coulomb'southward police force. Coulomb's law states that the electrical force between 2 charged objects is straight proportional to the production of the quantity of charge on the objects and inversely proportional to the square of the separation altitude between the 2 objects. In equation form, Coulomb's police can exist stated equally
where Q1 represents the quantity of charge on object 1 (in Coulombs), Q2 represents the quantity of charge on object 2 (in Coulombs), and d represents the distance of separation between the two objects (in meters). The symbol m is a proportionality constant known as the Coulomb'southward law constant. The value of this abiding is dependent upon the medium that the charged objects are immersed in. In the case of air, the value is approximately ix.0 x 109 Northward • thousand2 / C2. If the charged objects are present in water, the value of k can be reduced by as much as a factor of 80. It is worthwhile to point out that the units on k are such that when substituted into the equation the units on charge (Coulombs) and the units on distance (meters) will be canceled, leaving a Newton every bit the unit of forcefulness.
The Coulomb'southward law equation provides an accurate description of the force between ii objects whenever the objects human action every bit indicate charges . A charged conducting sphere interacts with other charged objects as though all of its charge were located at its center. While the charge is uniformly spread across the surface of the sphere, the middle of charge can be considered to exist the center of the sphere. The sphere acts as a bespeak accuse with its excess charge located at its center. Since Coulomb's law applies to point charges, the distance d in the equation is the distance between the centers of charge for both objects (not the distance between their nearest surfaces).
The symbols Q1 and Qtwo in the Coulomb's police equation represent the quantities of charge on the two interacting objects. Since an object tin be charged positively or negatively, these quantities are often expressed as "+" or "-" values. The sign on the accuse is simply representative of whether the object has an excess of electrons (a negatively charged object) or a shortage of electrons (a positively charged object). It might be tempting to utilize the "+" and "-" signs in the calculations of force. While the practice is not recommended, at that place is certainly no harm in doing so. When using the "+" and "-" signs in the adding of forcefulness, the result will be that a "-" value for strength is a sign of an bonny force and a "+" value for force signifies a repulsive forcefulness. Mathematically, the forcefulness value would be found to be positive when Q1 and Q2 are of like charge - either both "+" or both "-". And the strength value would exist found to exist negative when Q1 and Q2 are of opposite charge - one is "+" and the other is "-". This is consistent with the concept that oppositely charged objects have an bonny interaction and like charged objects accept a repulsive interaction. In the end, if you lot're thinking conceptually (and not merely mathematically), you would be very able to decide the nature of the force - attractive or repulsive - without the use of "+" and "-" signs in the equation.
Calculations Using Coulomb'south Law
In physics courses, Coulomb'south law is oft used equally a type of algebraic recipe to solve physics discussion bug. Three such examples are shown here.
Example A Suppose that two point charges, each with a charge of +i.00 Coulomb are separated past a distance of 1.00 meter. Determine the magnitude of the electrical force of repulsion between them.
This is non the most difficult mathematical problem that could exist selected. It certainly was non chosen for its mathematical rigor. The problem-solving strategy utilized here may seem unnecessary given the simplicity of the given values. Nonetheless, the strategy volition be used to illustrate its usefulness to any Coulomb'south police trouble.
The outset step of the strategy is the identification and listing of known information in variable form. Here nosotros know the charges of the two objects ( Qone and Q2 ) and the separation distance between them ( d ). The side by side step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the trouble requests information nigh the force. And so Felect is the unknown quantity. The results of the showtime two steps are shown in the table below.
| Given: Q1 = 1.00 C Q2 = 1.00 C d = one.00 m | Find: Felect = ??? |
The side by side and final step of the strategy involves substituting known values into the Coulomb's constabulary equation and using proper algebraic steps to solve for the unknown data. This step is shown beneath.
Felect = k • Q1 • Q2 / d2 Felect = (9.0 x 10nine N•10002/C2 ) • (1.00 C) • (ane.00 C) / (1.00 k)2
Felect = 9.0 x ten9 N
The strength of repulsion of two +1.00 Coulomb charges held ane.00 meter apart is 9 billion Newton. This is an incredibly large force that compares in magnitude to the weight of more 2000 jetliners.
This problem was chosen primarily for its conceptual bulletin. Objects only do not learn charges on the guild of ane.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 oftentimes used in front of the Coulomb as a unit of measurement 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 exchange into the Coulomb'south police equation. The following unit equivalencies will assist in such conversions.
1 Coulomb = 10vi microCoulomb 1 Coulomb = 109 nanoCoulomb
The problem-solving strategy used in Case A included 3 steps:
- Identify and list known data in variable course.
- List the unknown (or desired) data in variable grade.
- Substitute known values into the Coulomb's police equation and using proper algebraic steps to solve for the unknown information. (In some cases and for some students, it might exist easier to start practise the algebra using the variables and so perform the substitution as the last step.)
This same problem-solving strategy is demonstrated in Case B below.
Instance B Two balloons are charged with an identical quantity and type of charge: -6.25 nC. They are held apart at a separation distance of 61.seven cm. Determine the magnitude of the electric force of repulsion between them.
The problem states the value of Q1 and Qtwo . Since these values are expressed in units of nanoCoulombs (nC), the conversion to Coulombs must exist fabricated. The problem also states the separation altitude (d). Since distance is given in units of centimeters (cm), the conversion to meters must also be fabricated. 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 forcefulness (F). The results of the first two steps are shown in the table below.
| Given: Q1 = -6.25 nC = -6.25 x x-nine C Q2 = -6.25 nC = -half-dozen.25 x 10-9 C d = 61.7 cm = 0.617 g | Observe: Felect = ??? |
The final step of the strategy involves substituting known values into the Coulomb'southward law equation and using proper algebraic steps to solve for the unknown information. This commutation and algebra is shown below.
Felect = yard • Q1 • Q2 / d2 Felect = (9.0 x 109 N•mii/C2 ) • (vi.25 x 10-nine C) • (vi.25 x 10-9 C) / (0.617 m)2
Felect = nine.23 x 10-7 N
Note that the "-" sign was dropped from the Qi 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 strength 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 forcefulness is attractive (a "-" F value) or repulsive (a "+" F value).
Example C Two 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 2 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 likewise states the electric force (F). The unknown quantity is the separation distance (d). The results of the first two steps are shown in the table below.
| Given: Qi = +three.37 µC = +3.37 x 10-6 C Qii = -8.21 µC = -viii.21 x ten-vi C Felect = -0.0626 North (use a - force value since it is attractive) | Observe: d = ??? |
Every bit mentioned above, the use of the "+" and "-" signs is optional. However, if they are used, then they take to be used consistently for the Q values and the F values. Their use in the equation is illustrated in this problem.
The concluding step of the strategy involves substituting known values into the Coulomb'due south constabulary 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 = chiliad • Qane • Q2 / dtwo d2 • Felect = g • Qane • Q2
d2 = thousand • Q1 • Q2 / Felect
d = SQRT(k • Q1 • Q2 ) / Felect
d = SQRT [(9.0 x 109 N•one thousandtwo/C2 ) • (-8.21 x ten-half dozen C) • (+3.37 x 10-six C) / (-0.0626 N)]
d = Sqrt [ +iii.98 yard 2 ]
d = +1.99 m
Comparing Electrical and Gravitational Forces
Electric strength and gravitational force are the two non-contact forces discussed in The Physics Classroom tutorial. Coulomb's constabulary equation for electrical forcefulness bears a strong resemblance to Newton'south equation for universal gravitation.
The 2 equations have a very similar grade. Both equations testify an inverse square relationship between force and separation distance. And both equations bear witness that the force is proportional to the product of the quantity that causes the force - accuse in the instance of electrical strength and mass in the case of gravitational force. Yet there are some hit differences between these two forces. First, a comparing of the proportionality constants - chiliad versus G - reveals that the Coulomb'due south constabulary constant (yard) is significantly greater than Newton's universal gravitation constant (G). Subsequently a unit of charge will attract a unit of accuse with significantly more than strength than a unit of mass will attract a unit of mass. 2d, gravitational forces are only attractive; electrical forces tin can be either attractive or repulsive.
The inverse foursquare relationship betwixt force and distance that is woven into the equation is mutual to both non-contact forces. This human relationship highlights the importance of separation distance when it comes to the electric force betwixt charged objects. It is the focus of the next department of Lesson 3.
We Would Like to Suggest ...
Sometimes it isn't enough to just read about information technology. Y'all have to interact with information technology! And that'due south exactly what you exercise when you use one of The Physics Classroom's Interactives. Nosotros would like to suggest that you lot combine the reading of this page with the use of our Coulomb'southward Law Interactive. You can discover it in the Physics Interactives section of our website. The Coulomb'south Constabulary Interactive allows a learner to explore the effect of accuse and separation distance upon the amount of electric strength between two charged objects.
Check Your Understanding
Use your understanding to respond the following questions. When finished, click the button to view the answers.
ane. The Q in Coulomb's constabulary equation stands for the _____.
| a. mass of a charged object | b. # of excess electrons on the object |
| c. the current of a charged object | d. the altitude between charged objects |
| east. charge of a charged object | |
2. The symbol d in Coulomb'south law equation represents the distance from ___.
| a. A to B | b. A to D | c. B to C | d. B to D |
| due east. C to D | f. A to G | g. B to F | h. C to E |
3. Determine the electrical strength of attraction between two balloons with split up charges of +3.5 x 10-8 C and -two.9 x 10-eight C when separated a distance of 0.65 m.
4. Make up one's mind the electrical forcefulness of allure between two balloons that are charged with the reverse type of charge but the same quantity of accuse. The charge on the balloons is six.0 10 10-seven C and they are separated by a distance of 0.50 m.
5. Joann has rubbed a airship with wool to give information technology a charge of -1.0 x 10-vi C. She then acquires a plastic golf tube with a charge of +four.0 ten ten-vi C localized at a given position. She holds the location of charge on the plastic golf tube a altitude of fifty.0 cm above the balloon. Determine the electric force of allure betwixt the golf tube and the balloon.
6. A airship with a charge of 4.0 µC is held a altitude of 0.70 m from a second airship having the aforementioned charge. Calculate the magnitude of the repulsive force.
7. At what distance of separation must two 1.00-microCoulomb charges be positioned in lodge for the repulsive force between them to be equivalent to the weight (on Earth) of a 1.00-kg mass?
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