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Coulomb’s Law

Coulomb’s Law. Newton’s Law of Gravitation. The gravitational force between two masses, m1 & m2 is proportional to the product of the masses and inversely proportional to the square of the distance between them. F = G m 1 m 2 d 2. Newton’s Law of Gravitation. F = G m 1 m 2 d 2

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Coulomb’s Law

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  1. Coulomb’s Law

  2. Newton’s Law of Gravitation The gravitational force between two masses, m1 & m2 is proportional to the product of the masses and inversely proportional to the square of the distance between them. F = G m1m2 d2

  3. Newton’s Law of Gravitation F = G m1m2 d2 • G = the universal gravitational constant • m = mass • d = distance between the objects

  4. Coulomb’s Law Electrical force between two objects/charges has a similar inverse-square relationship with distance. Coulomb’s Law - for charged particles/objects, the force between the charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

  5. Coulomb’s Law • Coulomb’s Law gives you the magnitude of the force between the objects/charge. F = k q1q2 d2 d = distance between the charged particles q1 = the quantity of charge of one particle q2 = quantity of charge of the other particle k = the proportionality constant = 9.0 x 109 Nm2/C2 The SI unit of charge is the coulomb (C). 1C = 6.24 x 1018 electrons

  6. Remember that force is a vector, so when more than one charge exerts a force on another charge, the net force on that charge is the vector sum of the individual forces. Remember, too, that charges of the same sign exert repulsive forces on one another, while charges of opposite sign attract.

  7. Sample Problem What is the attractive electric force between a hydrogen atom’s proton and electron? qp= + 1.6 x 10-19 C qe= - 1.6 x 10-19 C r = 5.3 x 10-11 m (atomic radius) kC = 8.99 x 109 N-m2/C2

  8. Substitute and Solve Felect = 8.99 x 109 N.m2/C2(1.6 x 10-19 C) (1.6 x 10-19 C) (5.3 x 10-11 m)2 Felect = 8.19 x 10-8 N, attractive

  9. Example 2 One charge of 2.0 C is 1.5 m away from a – 3.0 C charge. Determine the force they exert on each other.

  10. Fe= k q1 q2 r2 Fe= (8.99e9 Nm2/C2)( 2.0 C)(−3.0 C) (1.52 m)2 Fe= −2.4 e10N The negative sign just means that one charge is positive, the other is negative, so there is an attractive force between them.

  11. Multiple Charges in One Dimension Things get a bit more interesting when you start to consider questions that have more than two charges. You will almost always deal with three charges in these linear problems. In the following example you have three charges lined up and are asked to calculate the net force acting on one of them. Do one step at a time, and then combine the answers at the end.

  12. Example 2: The following three charges are arranged as shown. Determine the net force acting on the charge on the far right (q3 = charge 3). q1= 1.5e-7 C q2 = - 2.3 e-7 C q3 =−3.5e-4 C 1.4 m 1.7 m

  13. Calculate the force between one pair of charges, then the next pair of charges, and so on until you have calculated all the possible combinations for that particular question. Remember, if you've calculated the force of q1 on q2, then you also know the force of q2 on q1 ... they're the same!

  14. Step 1: Calculate the force that charge 1 exerts on charge 3... It does NOT matter that there is another charge in between these two… ignore it! It will not effect the calculations that we are doing for these two. Notice that the total distance between charge 1 and 3 is 3.1 m , since we need to add 1.4 m and 1.7 m .

  15. F13 = k q1q3 r2 F13 = (8.99e9 Nm2/C2)( 1.5e-7 C )(−3.5e-4 C) (3.12 m)2 F13 = −4.9e-2N

  16. The negative sign just tells us the charges are opposite, so the force is attractive. Charge 1 is pulling charge 3 to the left, and vice versa. Do not automatically treat a negative answer as meaning “to the left” in this formula!!! Since all we care about is what is happening to charge 3, All we really need to know from this is that, charge 3 feels a pull towards the left of 4.9e-2 N.

  17. Step 2: Calculate the force that charge 2 exerts on charge 3... F23= k q2 q3 r2 F23 =8.99e9 Nm2/C2)(−2.3e-7 C)(−3.5e-4 C) (1.72 m)2 F23 = + 2.5e-1N The positive sign tells you that the charges are either both negative or both positive, so the force is repulsive. We know that charge 2 is pushing charge 3 to the right with a force of 2.5e-1 N.

  18. Step 3: Add you values to find the net force. Force acting on charge 3 is Fnet3 = F13 + F23 F13 = −4.9e-2N F23 =+2.5e-1N Fnet3 = −4.9e-2N + 2.5e-1N Fnet3 = 2.01e -1 N

  19. Class Problem Calculate the magnitude and determine the direction of the electric force between q1 and q2 and those between q1 and q3 per the diagram below: 60 cm 40 cm q1 q2 q3 +2 μC +4 μC - 3 μC

  20. Class Problem Calculate the magnitude and direction of Felect. on charge q1 in the example below Kc = 8.99 x 109 N-m2/C2 Felect. = kc q1q2/r2 3.0 cm 2.0 cm q2 q3 q1 + 6.0 C + 1.5 C - 2.0 C

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