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- 1. Coulomb's Law Equation 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. In equation form, Coulomb's law can be stated as 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 k is a proportionality constant known as the Coulomb's law constant. The value of this constant is dependent upon the medium that the charged objects are immersed in. In the case of air, the value is approximately 9.0 x 109 N • m2 / C2. Example A 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. Given: Q1 = 1.00 C Q2 = 1.00 C d = 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 • Q2 / d2 Felect = (9.0 x 109 N•m2/C2) • (1.00 C) • (1.00 C) / (1.00 m)2 Felect = 9.0 x 109 N Example 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
- 2. 61.7 cm. Determine the magnitude of the electrical force of repulsion between them. Given: Q1 = -6.25 nC = -6.25 x 10-9 C Q2 = -6.25 nC = -6.25 x 10-9 C d = 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 • Q2 / d2 Felect = (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 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 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 C Q2 = -8.21 µC = -8.21 x 10-6 C Felect = -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 • Q2 / d2 d2 • Felect = k • Q1 • Q2
- 3. d2 = k • Q1 • Q2 / 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 m2 ] d = +1.99 m Example: Find the force on the charge q2 in the diagram below due to the charges q1 and q2 q1 = 1 µC q2 = -2 µC q3 = 3 µC 0.1 m 0.15 m )righttheto(N.N.N.FFF )righttheto(N. )m.( )Cx)(Cx( )C/mNx.( r |q||q| kF )lefttheto(N. )m.( )Cx)(Cx( )C/mNx.( r |q||q| kF e e 604281 42 150 102103 10998 3 81 10 102101 10998 32122 2 66 229 2 32 2 32 2 66 229 2 12 21 12 Example:
- 4. An example Four charges are arranged in a square with sides of length 2.5 cm. The two charges in the top right and bottom left corners are +3.0 x 10-6 C. The charges in the other two corners are -3.0 x 10-6 C. What is the net force exerted on the charge in the top right corner by the other three charges? To solve any problem like this, the simplest thing to do is to draw a good diagram showing the forces acting on the charge. You should also let your diagram handle your signs for you. Force is a vector, and any time you have a minus sign associated with a vector all it does is tell you about the direction of the vector. If you have the arrows giving you the directionon your diagram, you can just drop any signs that come out of the equation for Coulomb's law. Consider the forces exerted on the charge in the top right by the other three: You have to be very careful to add these forces as vectors to get the net force. In this problem we can take advantage of the symmetry, and combine the forces from charges 2 and 4 into a force along the diagonal (opposite to the force from charge 3) of magnitude 183.1 N. When this is combined with the 64.7 N force in the opposite direction, the result is a net force of 118 N pointing along the diagonal of the square.
- 5. The symmetry here makes things a little easier. If it wasn't so symmetric, all you'd have to do is split the vectors up in to x and y components, add them to find the x and y components of the net force, and then calculate the magnitude and direction of the net force from the components. Example 16-4 in the textbook shows this process. Electric Field Electric Field A charged particle exerts a force on particles around it. We can call the influence of this force on surroundings as electric field. It can be also stated as electrical force per charge. Electric field is represented with E and Newton per coulomb is the unit of it. Electric field is a vector quantity. And it decreases with the increasing distance .k=9.109 Nm2 /C2 Electric field cannot be seen, but you can observe the effects of it on charged particles inside electric field. To find the electric field vector of a charge at one point, we assume that as if there is a +1 unit of charge there. If you want to find the total electric field of the charges more than one, you should find them one by one and add them using vector quantities. Electric Field Lines Motion path of the “+” charge in an electric field is called electric field line. Intensity of the lines shows the intensity of the electric field. Pictures given below show the drawings of field line of the positive charge and negative charge.
- 6. Electric field lines; Are perpendicular to the surfaces Never intercept If the electric field lines are parallel to each other, we call this regular electric field and it can be possible between two oppositely charged plates. E is constant within this plates and zero outside the plates. Pictures given below show the path of lines of two same charges and two opposite charges. Example:Find the electric field created by the charges A and B at point C in terms of k.q/d2?
- 7. Example:If the electric field at point A is zero, find the charge at point D in terms of q.

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