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706_PartUniversity Physics Solution

# 706_PartUniversity Physics Solution - Electric Potential...

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Electric Potential 23-29 (c) S ET U P : part (a): 0 0 ln ln 1 . 4 4 Q x a Q a V a x a x π π + = = + P P From Appendix B, 2 ln(1 ) / 2 ..., u u u + = so 2 2 ln(1 / ) / /2 a x a x a x + = and this becomes / a x when x is large. E XECUTE : Thus 0 0 . 4 4 Q a Q V a x a π π = P P For large x, V becomes the potential of a point charge. part (b): 2 2 2 2 0 0 ln ln 1 . 4 4 a a y Q Q a a V a y a y y π π + + = = + + P P From Appendix B, 2 2 2 2 1/ 2 2 2 1 / (1 / ) 1 /2 a y a y a y + = + = + + Thus 2 2 2 2 / 1 / 1 / / 2 1 / . a y a y a y a y a y + + + + + + And then using ln(1 ) u u + gives 0 0 0 ln(1 / ) . 4 4 4 Q Q a Q V a y a a y y π π π + = P P P E VALUATE : For large y, V becomes the potential of a point charge. 23.80. I DENTIFY : The potential at the surface of a uniformly charged sphere is kQ V R = . S ET U P : For a sphere, 3 4 3 V R π = . When the raindrops merge, the total charge and volume is conserved. E XECUTE : (a) 12 4 ( 1.20 10 C) 16.6 V 6.50 10 m kQ k V R × = = = − × . (b) The volume doubles, so the radius increases by the cube root of two: 4 3 new 2 8.19 10 m R R = = × and the new charge is 12 new 2 2.40 10 C. Q Q = = − × The new potential is 12 new new 4 new ( 2.40 10 C) 26.4 V 8.19 10 m kQ k V R × = = = − × . E VALUATE : The charge doubles but the radius also increases and the potential at the surface increases by only a factor of 2 / 3 1/ 3 2 2 2 = . 23.81. (a) I DENTIFY and S ET U P : The potential at the surface of a charged conducting sphere is given by Example 23.8: 0 1 . 4 q V R π = P For spheres A and B this gives 0 4 A A A Q V R π = P and 0 . 4 B B B Q V R π = P E XECUTE : A B V V = gives 0 0 /4 /4 A A B B Q R Q R π π = P P and / / . B A B A Q Q R R = And then 3 A B R R = implies / 1/3. B A Q Q = (b) I DENTIFY and S ET U P : The electric field at the surface of a charged conducting sphere is given in Example 22.5: 2 0 1 . 4 q E R π = P E XECUTE : For spheres A and B this gives 2 0 4 A A A Q E R π = P and 2 0 . 4 B B B Q E R π = P 2 2 2 0 2 0 4 / ( / ) (1/3)(3) 3. 4 B B A B A A B A B A Q E R Q Q R R E R Q π π = = = = ⎠⎝ P P E VALUATE : The sphere with the larger radius needs more net charge to produce the same potential. We can write / E V R = for a sphere, so with equal potentials the sphere with the smaller R has the larger V. 23.82. I DENTIFY : Apply conservation of energy, a a b b K U K U + = + . S ET U P : Assume the particles initially are far apart, so 0 a U = , The alpha particle has zero speed at the distance of closest approach, so 0 b K = . 19 1 eV 1.60 10 J = × . The alpha particle has charge 2 e + and the lead nucleus has charge 82 e + .

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