E2HWCH22 - Chapter 22 The Electric Field II Continuous...

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Unformatted text preview: Chapter 22 The Electric Field II: Continuous Charge Distributions 13 •• [SSM] A uniform line charge that has a linear charge density l equal to 3.5 nC/m is on the x axis between x = 0 and x = 5.0 m. ( a ) What is its total charge? Find the electric field on the x axis at ( b ) x = 6.0 m, ( c ) x = 9.0 m, and ( d ) x = 250 m. ( e ) Estimate the electric field at x = 250 m, using the approximation that the charge is a point charge on the x axis at x = 2.5 m, and compare your result with the result calculated in Part ( d ). (To do this you will need to assume that the values given in this problem statement are valid to more than two significant figures.) Is your approximate result greater or smaller than the exact result? Explain your answer. Picture the Problem We can use the definition of λ to find the total charge of the line of charge and the expression for the electric field on the axis of a finite line of charge to evaluate E x at the given locations along the x axis. In Part ( d ) we can apply Coulomb’s law for the electric field due to a point charge to approximate the electric field at x = 250 m. ( a ) Use the definition of linear charge density to express Q in terms of λ : ( 29 ( 29 nC 18 nC 17.5 m 5.0 nC/m 3.5 = = = = L Q λ Express the electric field on the axis of a finite line charge: ( 29 ( 29 L x x kQ x E x- = ( b ) Substitute numerical values and evaluate E x at x = 6.0 m: ( 29 ( 29 ( 29 ( 29 ( 29 N/C 26 m 5.0 m 6.0 m 6.0 nC 17.5 /C m N 10 8.988 m 6.0 2 2 9 =- ⋅ × = x E ( c ) Substitute numerical values and evaluate E x at x = 9.0 m: ( 29 ( 29 ( 29 ( 29 ( 29 N/C 4 . 4 m 5.0 m 9.0 m 9.0 nC 17.5 /C m N 10 8.988 m 9.0 2 2 9 =- ⋅ × = x E ( d ) Substitute numerical values and evaluate E x at x = 250 m: ( 29 ( 29 ( 29 ( 29 ( 29 mN/C 6 . 2 mN/C 56800 . 2 m 5.0 m 50 2 m 50 2 nC 17.5 /C m N 10 8.988 m 50 2 2 2 9 = =- ⋅ × = x E ( e ) Use Coulomb’s law for the electric field due to a point charge to obtain: ( 29 2 x kQ x E x = Substitute numerical values and evaluate E x (250 m): ( 29 ( 29 ( 29 ( 29 mN/C 6 . 2 mN/C 56774 . 2 m 2.5 m 250 nC 17.5 /C m N 10 8.988 m 250 2 2 2 9 = =- ⋅ × = x E This result is about 0.01% less than the exact value obtained in ( d ). This suggests that the line of charge is too long for its field at a distance of 250 m to be modeled exactly as that due to a point charge. 15 • A charge of 2.75 μ C is uniformly distributed on a ring of radius 8.5 cm. Find the electric field strength on the axis at distances of ( a ) 1.2 cm, ( b ) 3.6 cm, and ( c ) 4.0 m from the center of the ring. ( d ) Find the field strength at 4.0 m using the approximation that the ring is a point charge at the origin, and compare your results for Parts ( c ) and ( d ). Is your approximate result a good one? Explain your answer....
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E2HWCH22 - Chapter 22 The Electric Field II Continuous...

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