Electric Potential 23-29 (c) SET UP: part (a):00lnln 1.44QxaQaVaxaxππ+⎛⎞⎛⎞==+⎜⎟⎜⎟⎝⎠⎝⎠PPFrom Appendix B, 2ln(1)/ 2 ...,uuu+=−so 22ln(1/)//2a xa xax+=−and this becomes /a xwhen xis large. EXECUTE: Thus 00.44QaQVaxaππ⎛⎞→=⎜⎟⎝⎠PPFor large x,Vbecomes the potential of a point charge. part (b):222200lnln1.44aayQQaaVayayyππ⎡⎤⎛⎞⎛⎞++⎢⎥⎜⎟⎜⎟==++⎜⎟⎜⎟⎢⎥⎝⎠⎝⎠⎣⎦PPFrom Appendix B, 22221/ 2221/(1/)1/2ayayay+=+=++…Thus 2222/1/1// 21/.ayayayayay++→+++→+…And then using ln(1)uu+≈gives 000ln(1/).444QQaQVayaayyπππ⎛⎞→+→=⎜⎟⎝⎠PPPEVALUATE: 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 kQVR=. SET UP: For a sphere, 343VRπ=. When the raindrops merge, the total charge and volume is conserved. EXECUTE: (a)124( 1.2010C)16.6 V6.5010mkQkVR−−−×=== −×. (b) The volume doubles, so the radius increases by the cube root of two: 43new28.1910mRR−==×and the new charge is 12new22.4010C.QQ−== −×The new potential is 12newnew4new( 2.4010C)26.4 V8.1910mkQkVR−−−×=== −×. EVALUATE: The charge doubles but the radius also increases and the potential at the surface increases by only a factor of 2 / 31/ 3222=. 23.81. (a) IDENTIFY and SET UP: The potential at the surface of a charged conducting sphere is given by Example 23.8: 01.4qVRπ=PFor spheres Aand Bthis gives 04AAAQVRπ=Pand 0.4BBBQVRπ=PEXECUTE: ABVV=gives 00/4/4AABBQRQRππ=PPand //.BABAQQRR=And then 3ABRR=implies /1/3.BAQQ=(b) IDENTIFY and SET UP: The electric field at the surface of a charged conducting sphere is given in Example 22.5: 201.4qERπ=PEXECUTE: For spheres Aand Bthis gives 204AAAQERπ=Pand 20.4BBBQERπ=P2220204/(/)(1/3)(3)3.4BBABAABABAQERQQRRERQππ⎛⎞⎛⎞====⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠PPEVALUATE: The sphere with the larger radius needs more net charge to produce the same potential. We can write /EVR=for a sphere, so with equal potentials the sphere with the smaller Rhas the larger V.23.82. IDENTIFY: Apply conservation of energy, aabbKUKU+=+. SET UP:Assume the particles initially are far apart, so 0aU=, The alpha particle has zero speed at the distance of closest approach, so 0bK=. 191 eV1.6010J−=×. The alpha particle has charge 2e+and the lead nucleus has charge 82e+.
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