# chapt22 - Chapter 22 Electric Fields Problem 1 In Fig 22-30...

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Chapter 22: Electric Fields Problem 1 In Fig. 22-30 the electric field lines on the left have twice the separation as those on the right. (a) If the magnitude of the field at A is 40 N/C, what force acts on a proton at A? (b) What is the magnitude of the field at B? Answer (a) 6.40 × 10 -18 N. (b) 20 N/C. B A Figure 22-30 Solution (a) If E is the magnitude of the electric field at A then the magnitude of the force on the proton is F = qE, (q is the magnitude of the charge on proton) F = (1.60 × 10 -19 C)*(40 N/C) F = 6.40 × 10 N (b) Note that the field lines at B are about twice as far apart as at A. Since the density of the field lines represents the magnitude of the electric field this means that the magnitude of the field at B is about half of that at A or 20 N/C. Problem 3 What is the magnitude of a point charge whose electric field 50 cm away has the magnitude of 2.0 N/C? Answer 56 pC The magnitude of the electric field due to a point charge q at a point at a distance r from the point charge is E = 1 4 πε o q r 2 Therefore the magnitude of the charge is q = 4 πε o r 2 E = 0.50 m ( ) 2 2.0 N/C ( ) 8.99 × 10 9 N m 2 C 2 = 56 pC

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22-2 Chapter 22: Electric Fields Problem 8 (a) In Fig. 22-31, two point charges q 1 = -5q and q 2 = +2q are fixed to the x axis. (a) As a multiple of distance L, at what coordinate on the axis is the net electric field of the particles zero? (b) Sketch the electric field lines. L x y q 1 q 2 Figure 22-31 Answer (a) 1.72L to the right of q 2 . Solution (a) We can set up the equations for this problem if we will first determine possible regions in which the electric field could be zero. Referring to Diagram 8a, note first of all that the field cannot be zero accept along the line that runs through the two point charges. Furthermore by looking at the directions and relative sizes of the electric fields due to the two charges individually in three intervals along this line we see that any points where the field is zero must lie to the right of q 2 . L x y q 1 q 2 E q 1 q 2 E q 2 E q 2 E E q 1 E q 1 Diagram 8a Therefore if the field is zero it is at a point (x,0) where x > L. The total electric field at such a point is E tot = 1 4 πε o q 1 r 1 3 r 1 + 1 4 πε o q 2 r 2 3 r 2 where r 1 is the vector from q 1 to the point (x,0) and r 2 is the vector from q 2 to the point (x,0). Which means that r 1 = x i and r 2 = (x - L) i . Therefore if the total electric field is zero the previous equation for E tot becomes 0 = 1 4 πε o 5 q x 3 x i + 1 4 o 2 q x L ( ) 3 x L ( ) i or 0 = 5 x 2 + 2 x L ( ) 2 or 3x 2 – 10Lx + 5L 2 = 0.
Chapter 22: Electric Fields 22-3 Solving this quadratic equation for x yields x = 0.61L or x = 2.72L. According to our previous analysis x must be greater than L therefore we find that the field will be zero at a point 1.72L to the right of the q 2 . (b) A sketch of the field lines is shown in Diagram 8b. Notice in the sketch how the field is more strongly affected by the -5q charge. For quite a distance from the -5q charge the field still looks like that due to a single point charge as if the +2q charge were not present. In the vicinity of the +2q charge however the field deviates from the radial field

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## This note was uploaded on 08/03/2009 for the course PHYSICS 2049 taught by Professor C during the Spring '09 term at Vanderbilt.

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chapt22 - Chapter 22 Electric Fields Problem 1 In Fig 22-30...

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