Chapter 15 Coulomb’s Law Electrical Force Superposition

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

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...

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

What do we have so far...

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.

[F] = [k] [q 1 ] [q 2 ] / [r] 2 N = [k] C 2 /m 2 [k] = N m 2 /C 2 !

The value of k is determined experimentally to be… k = 8.99 X 10 9 N m 2 /C 2 We have now determined a quantitative expression for the electrostatic force! Let’s call it 9 X 10 9 N m 2 /C 2

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.

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!

Example: q1q1 q2q2 q3q3 q 1 = -1  C q 2 = -2  C q 3 = +1  C r 1 = 1 mr 2 =2 m What is the Total Force on q 2 ? r1r1 r2r2 y x 1) Start by examining the force exerted by q 1 on q 2. Points in the direction from 1 to 2! = (9 X 10 9 N m 2 / C 2 )(-1  C)(-2  C)/(1 m) 2 = +1.8 X N (i.e., in the +x direction). The plus sign indicates the force is repulsive.

Example (con’t): q1q1 q2q2 q3q3 q 1 = -1  C q 2 = -2  C q 3 = +1  C r 1 = 1 mr 2 =2 m What is the Total Force on q 2 ? r1r1 r2r2 y x 2) Then examine the force exerted by q 3 on q 2. Points in the direction from 3 to 2! = (9 X 10 9 N m 2 / C 2 )(+1  C)(-2  C)/(2 m) 2 = -4.5 X N The minus sign indicates the force is attractive…..therefore it’s in the +x direction.

Example (con’t): q1q1 q2q2 q3q3 q 1 = -1  C q 2 = -2  C q 3 = +1  C r 1 = 1 mr 2 =2 m What is the Total Force on q 2 ? r1r1 r2r2 y x 3) Finally, carefully add together the results. F 2 = F 12 + F 32 = 1.8 X N X N = 2.25 X N (i.e., in the +x direction).