可 6 Qoints fotel ceductros oupap ∽e thinki a V was (.43 i o -4: No■ cealitic ha&4s的 -3 Bandom sign ercocS d 可 PorA13 Y Reia+已e anctions -5 甲a5als,0cwc-y4hn 一 is a vecfo - Doin o。hycembis4iorq）p4ir :orin 士cinon0mth中以 0 -1 3■ onS w:☒ 的9 ecrocs +24W8 o 49f西r 2.OUerall s的7 6
6
Problem 2(25 points):answers without work shown will not be given any credit. A stick with length L has a positive charge O uniformly distributed on it.It lies along the x-axis between the points x=0 and x=L.A point-like object with identical positive charge O lies on the x-axis at the point x=2L. uniformly charged stick point-like object of length L and charge O with charge O x=0 x=L x=a x=2L You may find the following integrals useful: 人 =-ln(a-x) x dx J (a-x)(a-x) jeya严 dx 1 xdx 1 62+y Part(a) Let x=a be the point on the x-axis between the two objects where the electric field is zero.Find a.Express your answer in terms of k=1/4,,and L as needed. Part(b) Choose the zero point for the electric potential to be at infinity V()=0.Determine the electric potential at the point x=a.You may express your answer in terms of any of the quantities mentioned in part (a),whether or not you answered part (a);in particular in terms of k=1/4n,,L,and a as needed. 7
7 Problem 2 (25 points): answers without work shown will not be given any credit. A stick with length L has a positive charge Q uniformly distributed on it. It lies along the x -axis between the points x = 0 and x = L . A point-like object with identical positive charge Q lies on the x -axis at the point x = 2L . You may find the following integrals useful: dx a − x = −ln(a − x) ∫ dx (a − x) 2 = 1 (a − x) ∫ dx (x 2 + y 2 ) 3/2 = 1 y 2 x (x 2 + y 2 ) ∫ 1/2 xdx (x 2 + y 2 ) 3/2 = − 1 (x 2 + y 2 ) ∫ 1/2 Part (a) Let x = a be the point on the x -axis between the two objects where the electric field is zero. Find a . Express your answer in terms of k = 1/ 4πε 0 , Q , and L as needed. Part (b) Choose the zero point for the electric potential to be at infinity V (∞) = 0 . Determine the electric potential at the point x = a . You may express your answer in terms of any of the quantities mentioned in part (a), whether or not you answered part (a); in particular in terms of k = 1/ 4πε 0 , Q , L , and a as needed
ccblcm 2 (c) dg=sax! P X=a X=L Erod()=k k L X=o X-。(a-x2 yL CR- :e()9 ←2L-&7 K=( Ee(m:k&,(-) 2L-a)2 E(P)=Er(p)+E。（)-⊙ (e(）经)父à 习 t(i)ca 之 (a-L)a BL-a →@L-)=@-1)a> 42-4La+a2-a2-4L今4L2=3L之 =4L 8
8
(6) ←-x73 a口 +X Vrod los)=o w dx X=a dj-cdx! x上L x-L e-kcIn@~x） rod s,P XEo la-x') (=c ):(g)m(是 +& G X二气 X=AL Vs(o)=0 ∠2L-47 Vo(3)-Vs(0)-KQ Vc(?)-kc aL-a L-a V)-V(nv()kei(t ke 2L-& ena二4 Ym)=竖n(4)+是e=((4)+) 9
9