706_PartUniversity Physics Solution

706_PartUniversity Physics Solution - Electric Potential...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Electric Potential 23-29 (c) SET UP: part (a): 00 ln ln 1 . 44 Qx aQ a V ax a x ππ + ⎛⎞ == + ⎜⎟ ⎝⎠ PP From Appendix B, 2 ln(1 ) /2. .., uuu +=− so 22 ln(1 / ) / / 2 ax ax a x += and this becomes / when x is large. EXECUTE: Thus . Qa Q V a →= For large x, V becomes the potential of a point charge. part (b): 2 2 ln ln 1 . aay QQ a a V ay a y y ⎡⎤ ⎛⎞⎛⎞ ++ ⎢⎥ ⎜⎟⎜⎟ + + ⎣⎦ From Appendix B, 1 / 2 2 2 1/ ( )1/ 2 a y + = + + Thus 2 2 /1 / 1 / / 2 1 / . ay a y + + →+ + + →+ And then using ln(1 ) uu + gives 0 / ) . 4 a Q Va y aa y y π = P EVALUATE: For large y, V becomes the potential of a point charge. 23.80. IDENTIFY: The potential at the surface of a uniformly charged sphere is kQ V R = . SET UP: For a sphere, 3 4 3 VR = . When the raindrops merge, the total charge and volume is conserved. EXECUTE: (a) 12 4 (1 . 2 01 0 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 28 . 1 9 1 0 m RR × and the new charge is 12 new . 4 0 1 0 C . × The new potential is 12 new new 4 new (2 . 4 ) 26.4 V 8.19 10 m kQ k V R = × . EVALUATE: 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) IDENTIFY and SET UP: The potential at the surface of a charged conducting sphere is given by Example 23.8: 0 1 . 4 q V R = P For spheres A and B this gives 0 4 A A A Q V R = P and 0 . 4 B B B Q V R = P EXECUTE: AB VV = gives /4 AA BB QR = and / / . B A QQ RR = And then 3 R R = implies / 3 . BA = (b) IDENTIFY and SET UP: The electric field at the surface of a charged conducting sphere is given in Example 22.5: 2 0 1 . 4 q E R = P EXECUTE: For spheres A and B this gives 2 0 4 A A A Q E R = P and 2 0 . 4 B B B Q E R = P 2 0 2 0 4 /( / )( 1 / 3 ) ( 3 )3 . 4 B BA AB A Q ER Q = = P P EVALUATE: The sphere with the larger radius needs more net charge to produce the same potential. We can write / E VR = for a sphere, so with equal potentials the sphere with the smaller R has the larger V. 23.82. IDENTIFY: Apply conservation of energy, aa bb K UKU +=+ . SET UP: 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 + .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
23-30 Chapter 23 EXECUTE: Set the alpha particle’s kinetic energy equal to its potential energy: ab K U = gives (2 )(82 ) 11.0 MeV ke e r = and 19 2 14 61 9 (164)(1.60 10 C) 2.15 10 m ( 1 1 . 01 0eV ) ( 1 . 6 0 JeV ) k r × == × ×× . EVALUATE: The calculation assumes that at the distance of closest approach the alpha particle is outside the radius of the lead nucleus. 23.83. IDENTIFY and SET UP: The potential at the surface is given by Example 23.8 and the electric field at the surface is given by Example 22.5. The charge initially on sphere 1 spreads between the two spheres such as to bring them to the same potential. EXECUTE: (a) 1 1 2 1 , 4 Q E R π = P 1 11 1 1 4 Q VR E R P (b) Two conditions must be met: 1) Let q 1 and q 2 be the final potentials of each sphere. Then 12 1 qqQ + = (charge conservation) 2) Let V 1 and V 2 be the final potentials of each sphere. All points of a conductor are at the same potential, so 12 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

Page1 / 5

706_PartUniversity Physics Solution - Electric Potential...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online