# E2HWCH21 - Chapter 21 The Electric Field I Discrete Charge...

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Chapter 21 The Electric Field I: Discrete Charge Distributions 22 A charge equal to the charge of Avogadro’s number of protons ( N A = 6.02 × 10 23 ) is called a faraday . Calculate the number of coulombs in a faraday. Picture the Problem One faraday = N A e . We can use this definition to find the number of coulombs in a faraday. Use the definition of a faraday to calculate the number of coulombs in a faraday: ( 29 ( 29 C 10 63 . 9 C/electron 10 602 . 1 electrons 10 02 . 6 faraday 1 4 19 23 A × = × × = = - e N 26 A point charge q 1 = 4.0 μ C is at the origin and a point charge q 2 = 6.0 C is on the x -axis at x = 3.0 m. ( a ) Find the electric force on charge q 2 . ( b ) Find the electric force on q 1 . ( c ) How would your answers for Parts ( a ) and ( b ) differ if q 2 were –6.0 C? Picture the Problem We can find the electric forces the two charges exert on each by applying Coulomb’s law and Newton’s 3 rd law. Note that i r ˆ ˆ 2 , 1 = because the vector pointing from q 1 to q 2 is in the positive x direction. The diagram shows the situation for Parts ( a ) and ( b ). ( a ) Use Coulomb’s law to express the force that q 1 exerts on q 2 : 2 , 1 2 2 , 1 2 1 2 , 1 ˆ r F r q kq = Substitute numerical values and evaluate 2 , 1 F : ( 29 ( 29 ( 29 ( 29 ( 29 i i F ˆ mN 24 ˆ m 3.0 μC 6.0 μC 4.0 /C m N 10 8.988 2 2 2 9 2 , 1 = × = ( b ) Because these are action-and- reaction forces, we can apply Newton’s 3 rd law to obtain: ( 29 i F F ˆ mN 24 2 , 1 1 , 2 - = - = ( c ) If q 2 is - 6.0 C, the force between q 1 and q 2 is attractive and both force vectors are reversed:

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( 29 ( 29 ( 29 ( 29 ( 29 i i F ˆ mN 24 ˆ m 3.0 μC 6.0 μC 4.0 /C m N 10 8.988 2 2 2 9 2 , 1 - = - × = and ( 29 i F F ˆ mN 24 2 , 1 1 , 2 = - = 38 Two point charges, each +4.0 μ C, are on the x axis; one point charge is at the origin and the other is at x = 8.0 m. Find the electric field on the x axis at ( a ) x = –2.0 m, ( b ) x = 2.0 m, ( c ) x = 6.0 m, and ( d ) x = 10 m. ( e ) At what point on the x axis is the electric field zero? ( f ) Sketch E x versus x for –3.0 m < x < 11 m. Picture the Problem Let q represent the point charges of +4.0 C and use Coulomb’s law for E due to a point charge and the principle of superposition for fields to find the electric field at the locations specified. Noting that q 1 = q 2 , use Coulomb’s law and the principle of superposition to express the electric field due to the given charges at point P a distance x from the origin: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 - + = - + = - + = + = P , 2 P , 2 2 P , 2 P , 2 1 P , 2 2 P , 2 1 2 1 2 1 2 1 2 1 ˆ m 0 . 8 1 ˆ 1 /C m kN 36 ˆ m 0 . 8 1 ˆ 1 ˆ m 0 . 8 ˆ q q q q q q q q r x r x r x r x kq r x kq r x kq x E x E x E ( a ) Apply this equation to the point at x = - 2.0 m: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 i i i E ˆ kN/C 4 . 9 ˆ m 10 1 ˆ m 0 . 2
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E2HWCH21 - Chapter 21 The Electric Field I Discrete Charge...

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