Chapter22

# Chapter22 - 1(a We note that the electric field points...

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1. (a) We note that the electric field points leftward at both points. Using GG Fq E = 0 , and orienting our x axis rightward (so ˆ i points right in the figure), we find () 19 18 N ˆˆ 1.6 10 C 40 i ( 6.4 10 N)i C F −− ⎛⎞ =+ × =− × ⎜⎟ ⎝⎠ G which means the magnitude of the force on the proton is 6.4 × 10 –18 N and its direction ˆ (i ) is leftward. (b) As the discussion in §22-2 makes clear, the field strength is proportional to the crowdedness of the field lines. It is seen that the lines are twice as crowded at A than at B , so we conclude that E A = 2 E B . Thus, E B = 20 N/C.

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2. We note that the symbol q 2 is used in the problem statement to mean the absolute value of the negative charge which resides on the larger shell. The following sketch is for 12 qq = . The following two sketches are for the cases q 1 > q 2 (left figure) and q 1 < q 2 (right figure).
3. Since the magnitude of the electric field produced by a point charge q is given by 2 0 || / 4 Eq r πε = , where r is the distance from the charge to the point where the field has magnitude E , the magnitude of the charge is () 2 21 1 0 92 2 0.50m 2.0 N C 45 . 6 1 0 C . 8.99 10 N m C qr E ε = = × ×⋅

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4. We find the charge magnitude | q | from E = | q |/4 πε 0 r 2 : () ( ) 2 21 0 0 92 2 1.00 N C 1.00m 41 . 1 1 1 0 C . 8.99 10 N m C qE r = = × ×⋅ ε
5. Since the charge is uniformly distributed throughout a sphere, the electric field at the surface is exactly the same as it would be if the charge were all at the center. That is, the magnitude of the field is E q R = 4 0 2 π ε where q is the magnitude of the total charge and R is the sphere radius. (a) The magnitude of the total charge is Ze , so E Ze R == ×⋅ × × 4 8 99 10 94 160 10 664 10 307 10 0 2 92 2 1 9 15 2 21 π .. . Nm C C m NC ch bg (b) The field is normal to the surface and since the charge is positive, it points outward from the surface.

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6. With x 1 = 6.00 cm and x 2 = 21.00 cm, the point midway between the two charges is located at x = 13.5 cm. The values of the charge are q 1 = – q 2 = – 2.00 × 10 –7 C, and the magnitudes and directions of the individual fields are given by: () 92 2 7 5 1 1 2 2 01 2 7 5 2 2 2 2 02 || (8.99 10 N m C )| 2.00 10 C| ˆˆ ˆ i i (3.196 10 N C)i 4( ) 0.135 m 0.060 m (8.99 10 N m C )(2.00 10 C) ˆ i i (3.196 10 N C)i ) 0.135 m 0.210 m q E xx q E πε ×⋅ × =− × × × G G Thus, the net electric field is 5 net 1 2 ˆ (6.39 10 N C)i EE E =+= × GG G
Let x be the coordinate of P , the point where the field vanishes. Then, the total electric field at P is given by () 21 2 2 02 1 || 1 4() qq E xx πε ⎛⎞ =− ⎜⎟ ⎝⎠ . If the field is to vanish, then 2 2 2 22 2 11 ||( ) . | | q x x q =⇒ = −− Taking the square root of both sides, noting that | q 2 |/| q 1 | = 4, we obtain 70 cm 2.0 20 cm x x = ± . Choosing –2.0 for consistency, the value of x is found to be x = 30 cm. 7. At points between the charges, the individual electric fields are in the same direction and do not cancel. Since charge q 2 = 4.00 q 1 located at x 2 = 70 cm has a greater magnitude than q 1 = 2.1 × 10 8 C located at x 1 = 20 cm, a point of zero field must be closer to q 1 than to q 2 . It must be to the left of q 1 .

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Thus, we obtain 2.72 12 5 L x L
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## This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter22 - 1(a We note that the electric field points...

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