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Chapter 15 Coulomb’s Law Electrical Force Superposition

Chapter 15 Coulomb’s Law Electrical Force Superposition. Lecture 14. 10 February 1999 Wednesday. Physics 112. Last time: two types of charge Positive and Negative Quantized in units of +/- 1, +/- 2, etc.

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Chapter 15 Coulomb’s Law Electrical Force Superposition

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  1. Chapter 15 Coulomb’s Law Electrical Force Superposition Lecture 14 10 February 1999 Wednesday Physics 112

  2. Last time: two types of charge Positive and Negative Quantized in units of +/- 1, +/- 2, etc. The mks unit for charge is the Coulomb, named after Charles Coulomb in honor of his many contributions in the field of electricity. 1e = 1.6 X 10-19 C Remember to Convert!

  3. Quantifying the Electrical Force Joseph Priestly and Charles Coulomb set out to quantify the electrical force in the late 1700’s. Let’s see if we can figure it out an expression for the electric force ourselves... On What Does the Electric Force Depend?

  4. F = ??? Distance Charge on Body 1 Charge on Body 2 Anything else? 1/r2 (units of 1/m2) q1 (units of C) q2 (units of C) Just a constant of proportionality… Let’s call it k.

  5. What do we have so far... Coulomb's Law

  6. Note on Notation! The arrow indicates that F is a vector quantity (i.e., to specify F, you need both magnitude and direction)! Indicates a direction radially away from the center of our coordinate system. It could be the x-direction, the y-direction, or something in between.

  7. Units! [F] = [k] [q1] [q2] / [r]2 N = [k] C2/m2 [k] = N m2/C2!

  8. The value of k is determined experimentally to be… k = 8.99 X 109 N m2/C2 Let’s call it 9 X 109 N m2/C2 We have now determined a quantitative expression for the electrostatic force!

  9. Notes on Coulomb’s Law: Applies only to point charges, particles or spherical charge distributions. Obeys Newton’s 3rd Law. The electrical force, like gravity, is a “field” force…that is, a force is exerted at a distance despite lack of physical contact.

  10. The electrostatic force obeys the Superposition Principle This implies that to solve problems with multiple charges, we may consider each two charge system separately and combine the results at the end. Remember, force is a vector quantity, so you must use vector addition!

  11. Points in the direction from 1 to 2! = (9 X 109 N m2 / C2)(-1 mC)(-2 mC)/(1 m)2 = +1.8 X 10-2 N(i.e.,in the +x direction). The plus sign indicates the force is repulsive. Example: y q1 q2 q3 r1 r2 x q1 = -1 mC q2 = -2 mC q3 = +1 mC r1 = 1 m r2=2 m What is the Total Force on q2? 1) Start by examining the force exerted by q1 on q2.

  12. Points in the direction from 3 to 2! = (9 X 109 N m2 / C2)(+1 mC)(-2 mC)/(2 m)2 = -4.5 X 10-3 NThe minus sign indicates the force is attractive…..therefore it’sin the +x direction. Example (con’t): y q1 q2 q3 r1 r2 x q1 = -1 mC q2 = -2 mC q3 = +1 mC r1 = 1 m r2=2 m What is the Total Force on q2? 2) Then examine the force exerted by q3 on q2.

  13. Example (con’t): y q1 q2 q3 r1 r2 x q1 = -1 mC q2 = -2 mC q3 = +1 mC r1 = 1 m r2=2 m What is the Total Force on q2? 3) Finally, carefully add together the results. F2 = F12 + F32 = 1.8 X 10-2 N + 4.5 X 10-3 N = 2.25 X 10-2 N (i.e., in the +x direction).

  14. Concept Quiz! Electrical Forces

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