1. Chapter 16 Section 2 Electric force

2. Objectives  Calculate electric force using Coulomb’s law.  Compare electric force with gravitational force.  Apply the superposition principle to find the resultant force on a charge and to find the position at which the net force on a charge is zero.

3. Coulomb's Law  Two charges near one another exert a force on one another called the electric force.  Coulomb’s law states that the electric force is proportional to the magnitude of each charge and inversely proportional to the square of the distance between them.

4. Coulomb’s Law  The quantitative expression for the effect of these three variables on electric force is known as Coulomb's law. Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

5. Coulomb’s Law  The resultant force on a charge is the vector sum of the individual forces on that charge.  Adding forces this way is an example of the principle of superposition.  When a body is in equilibrium, the net external force acting on that body is zero.

6. Sample Problem The Superposition Principle Consider three point charges at the corners of a triangle, as shown at right, where q 1 = 6.00  10 –9 C, q 2 = –2.00  10 –9 C, and q 3 = 5.00  10 –9 C. Find the magnitude and direction of the resultant force on q 3.

7. solution The Superposition Principle 1. Define the problem, and identify the known variables. Given: q 1 = +6.00  10 –9 Cr 2,1 = 3.00 m q 2 = –2.00  10 –9 Cr 3,2 = 4.00 m q 3 = +5.00  10 –9 Cr 3,1 = 5.00 m  = 37.0º Unknown:F 3,tot = ? Diagram:

8. solution Tip: According to the superposition principle, the resultant force on the charge q 3 is the vector sum of the forces exerted by q 1 and q 2 on q 3. First, find the force exerted on q 3 by each, and then add these two forces together vectorially to get the resultant force on q 3. 2. Determine the direction of the forces by analyzing the charges. The force F 3,1 is repulsive because q 1 and q 3 have the same sign. The force F 3,2 is attractive because q 2 and q 3 have opposite signs.

9. solution  3. Calculate the magnitudes of the forces with Coulomb’s law.

10. solution Find the x and y components of each force. At this point, the direction each component must be taken into account. F 3,1 : F x = (F 3,1 )(cos 37.0º) = (1.08  10 –8 N)(cos 37.0º) F x = 8.63  10 –9 N F y = (F 3,1 )(sin 37.0º) = (1.08  10 –8 N)(sin 37.0º) F y = 6.50  10 –9 N F 3,2 :F x = –F 3,2 = –5.62  10 –9 N F y = 0 N

11. solution  6. Use the Pythagorean theorem to find the magnitude of the resultant force.

12. solution Use a suitable trigonometric function to find the direction of the resultant force. In this case, you can use the inverse tangent function

13. Example  Suppose 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.

14. solution Given: Q 1 = 1.00 C Q 2 = 1.00 C d = 1.00 m Find: F elect = ??? F elect = k Q 1 Q 2 / d 2 F elect = (9.0 x 10 9 Nm 2 /C 2 ) (1.00 C) (1.00 C) / (1.00 m) 2 F elect = 9.0 x 10 9 N 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 (Q 1 and Q 2 ) 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 F elect is the unknown quantity. The results of the first two steps are shown in the table below

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

16. solution  The problem states the value of Q 1 and Q 2. 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: Q 1 = -6.25 nC = -6.25 x 10 -9 C Q 2 = - 6.25 nC = -6.25 x 10 -9 C d = 61.7 cm = 0.617 m Find: F elect = ???

17. solution  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.  F elect = k Q 1 Q 2 / d 2 F elect = (9.0 x 10 9 Nm 2 /C 2 ) (6.25 x 10 -9 C) (6.25 x 10 -9 C) / (0.617 m) 2  F elect = 9.23 x 10 -7 N

18. Example  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 two balloons.

19. solution Given: Q 1 = +3.37 µC = +3.37 x 10 -6 C Q 2 = - 8.21 µC = -8.21 x 10 -6 C F elect = -0.0626 N (use a - force value since it is attractive) Find: d = ??? The problem states the value of Q 1 and Q 2. 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 F elect = k Q 1 Q 2 / d 2 d 2 F elect = k Q 1 Q 2 d 2 = k Q 1 Q 2 / F elect d = SQRT(k Q 1 Q 2 ) / F elect d = SQRT [(9.0 x 10 9 Nm 2 /C 2 ) (-8.21 x 10 -6 C) (+3.37 x 10 -6 C) / (-0.0626 N)] d = Sqrt [ +3.98 m 2 ] d = +1.99 m

20. Coulomb’s Law  The Coulomb force is a field force.  A field force is a force that is exerted by one object on another even though there is no physical contact between the two objects.

21. Watch videos Lets watch videos http://www.youtube.com/watch?v=tDONPAqvnqk&feature=player_detailpage http://www.youtube.com/watch?v=FL3n8_SjanI&feature=player_detailpage http://app.discoveryeducation.com/player/view/assetGuid/8E69A6E4-ACD5- 4B13-800E-417CEBFCA876

22. Student guided practice  Let’s do worksheet

23. closure  Today we learned about electric force and coulomb’s law  Next class we are going to learn about electric fields