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Coulomb’s Law Electricity Equations are Similar to Gravity m1 Fg = m2 g1 q1 m2 Fg = G m2 g = G g1 = G m1m2 r m1 r 2 2 m1m2 r 2 q2 FE = q2 E1 q2 E = k E1 = k FE = k q1q2 r q1 r 2 2 q1q2 r 2 Solving for the Field m1 P mnew q1 qP new m g =G 2 r q E=k 2 r Fg = mnew g FE = qnew E Often problems begin by finding the field strength at point P . Then a new mass/charge appears at point P . Typically the problem then asks for the force on the new mass/charge caused by the field of the original mass/charge. Fields Near Point Masses and Point Charges A point mass/charge is an object where the field acts as though it is coming from a single point at the center of the object. Spheres M + − Note: The fields around the charges only look like this when they are very very very far away from any other charge. We will soon see that the field of a proton and electron next to each other looks different. Example 2 A spherical conductor has a charge of 4 μC . a. How are the charges distributed on the conductor? Excess charge piles up on the surface of conductors. b. Determine the magnitude of the electric field at a point P , located 30 cm to the right of the charge. ( q E = k 2 = 9 ´ 109 r () 4 ´ 10-6 (0.30) 2 4 μC P E ) = 4.0 ´ 10 N C 5 c. State the direction of the electric field. 1st Method: Charge is positive, and positive means away. field is +x 2nd Method: Imagine a positive test charge at point P. Now ask which direction the test charge would move. It will be repelled by the positive charge, so it will move right, or +x . Example 2 A spherical conductor has a charge of 4 μC . A −0.5 μC charge is inserted at point P . P E −0.5 μC 4 μC FE d. Determine the magnitude of force on the point charge ( )( ) FE = qE = 0.5 ´ 10-6 4.0 ´ 105 = 0.20 N e. Determine the direction of force on the point charge. 1st Method: The new charge is negative, and negative charges move opposite the electric field ( − means opposite or 180o). The field is +x , so the force has to be −x . 2nd Method: If an object is released from rest it will accelerate in the direction of force. Opposite charges atttract. When the new charge is released it will be attracted to the positive charge and move to the left, −x . Coulomb’s Law and the Force of Electricity Electricity is a field force similar to gravity. The equations follow the same patterns and they are worth comparing in order to reinforce each other. Newton’s Law of Universal Gravitation Coulomb’s Law m1m2 Fg = -G 2 r q1q2 FE = k 2 r G = 6.67 ´10 -11 k = 9 ´10 9 Coulomb’s Law and the Force of Electricity q1q2 FE = k 2 r k = 9 ´109 This version of Coulomb’s Law and the constant k are used in most situations in this course. The formal variant of the formula has certain advantages. 1 q1q2 FE = 4p Î0 r 2 Î0 = 8.85 ´ 10 Where the constant is the permittivity of free space 1 1 9 k= = = 9 ´ 10 -12 4p Î0 4p 8.85 ´ 10 ( ) -12 Example 1 Fg = -G ( Determine the force of gravity and the force of electricity between a +1.0 C charge with a mass of 1.0 × 10−8 kg and −1.0 C charge with a mass of 5.7 × 10−12 kg separated by 1 m . m1m2 r FE = k 2 Fg = 6.67 ´ 10 () 1.0 ´ 10 ) (5.7 ´ 10 ) -8 -11 Fg = 3.8 ´ 10-30 N (1) -12 2 FE q1q2 r2 1.0 ) (1.0 ) ( = ( 9 ´ 10 ) (1) 9 2 FE = 9.0 ´ 109 N Electricity is many time stronger than gravity. As a result, gravity is often negligible compared to electricity. As a rule if you cannot see the charged object with you then its mass is too small to be significant when compared with gravity. Newton’s 3rd Law Revisited FE = k q1q2 r 2 Is the force calculated by the equations the force of the proton pulling on the electron, or the electron pulling on the proton ? It is both. Newton’s 3rd Law states that when two objects interact there is an equal and opposite force between them. The electron may be many time smaller than the proton, but it pulls on the proton with equal force. The signs in the equations FE = k q1q2 r 2 Plugging in positive and negative values for charge q is not necessary. If you do there are two possible outcomes. A Positive answer means: Away, Repel A Negative answer means: Toward, Attract We already know that opposite charges attract, and that like charges repel. We can look at the sign of the charge in the diagram to decide which way charges will move. We don’t need the sign from the equation to tell us what is already obvious. However, calculus and graphing are done very formally and the minus signs are important. The signs in the equations + − +m1 + m2 -Fg = -G r2 − + +q1 - q2 -FE = k r2 Negative = Toward Negative = Toward + + + +m1 + m2 -Fg = -G r2 + +q1 + q2 +FE = k r2 Negative = Toward Positive = Away − − − +m1 + m2 -Fg = -G r2 Negative = Toward − + FE = k -q1 - q2 r2 Positive = Away Inverse Square Law m1m2 Fg = -G 2 r q1q2 FE = k 2 r Both the Universal Law of Gravitation and Coulomb’s Law have distance between the objects in the denominator Inverse and the distance is squared Square If distance of separation changes, then force also changes by a factor equal to the inverse square of change in distance. If the separation distance doubles, then force is ¼ of its original strength. If the separation distance is cut in half, then force is 4 times greater than before. Motion Perpendicular to a Radial Field The motion is circular It is the same for gravity and electricity In electricity the particles must have opposite charge. m M v0 − + v0 + v0 − In electricity this is a simplistic way of describing the motion of an electron around a proton in a hydrogen atom. Motion Perpendicular to a Radial Field Solving for any circular motion force problem requires the use of the circular motion sum of forces. Gravity Electricity Fc = Fg Fc = FE v Mm m =G 2 r r v Qq m =k 2 r r 2 GM v= r 2 v= kQq mr Example 3 An electron orbits a proton at a radius of 1.2 × 10−10 m. Determine the speed of the orbiting electron. Assume a circular orbit. Fc = FE v Qq m =k 2 r r 2 v= kQq mr v= ( )( )( 9 ´ 109 1.6 ´ 10-19 1.6 ´ 10-19 (9.11 ´ 10 ) (1.2 ´ 10 ) -31 v = 1.45 ´ 10 m s 6 -10 )