PHYS1252, Spring 2018, Exam#2 Solution Problem I: Multiple ...

1 PHYS1252, Spring 2018, Exam#2 Solution Problem I: Multiple - choice Exam #1A Exam #1B Exam #1C Exam #1D Exam #1E Exam #1F 1 A A A A A A 2 B A A A B A 3 D A D D D B 4 A E C C A E PHYS1252 Exam #2, Sp18 Problems II and IV: Force and Electric Field from Point Charges Solution & Grading Key The following detailed worked out Solution and Grading key uses the input parameter values of an analogous Problem from an older practice exam, shown on the next page. exams and IV, 2B P. II and IV, 2C P. II and IV, 2D P. II and IV, 2E P.

2 II and IV, 2F P. II and IV are solved analogously with intermediate and final results summarized in a summary table, following the detailed Solution and grading key. All required drawings are shown on pages following the summary table.. Practice Problem Input Parameters Used for Detailed Solution and Grading Key .. Physics 1252 Section 27180 Exam #2 AThu, 23 March 2017 Name:II. Force and Electric Field from Point Charges(40 points)Two point charges,Q1= andQ2=+ , are positioned atx1= and atx2=+ , respectively, on each part below, (a), (b), (c), you have to make a separate clean,BIGdrawing to getfull credit for the Problem , see next page.

3 (a) A small test charge,q=+118 C, is placed on thex-axis atxq= 0m. Find the net forceexerted onq, jointly byQ1andQ2. State all three components of the force vector,~F=(Fx,Fy,Fz), and state the magnitude of the force,F=|~F|. [20 points](b) Find the electric field generated jointly byQ1andQ2at a pointP, located atxP= on thex-axis. State all three components of the electric field vector,~E=(Ex,Ey,Ez), and state its magnitude,E=|~E|. [15 points](c) Repeat part (a), ifQ1andQ2are moved to the new locations (x1,y1,z1)=( , ,0)mand (x2,y2,z2)=(0, ,0)m, in thex-y-plane, with all other input parameters re-maining unchanged.

4 [5 points]Copyrightc 2017 University of 1252 Section 27180 Exam #2 AThu, 23 March 2017 Name:IV. Electric Field Vector in thex-y-Plane(10 points, Bonus)Repeat part (b) of Problem II, ifQ1andQ2are moved to the new locations (x1,y1,z1)=( , ,0)m and (x2,y2,z2)=(0, ,0)m, respectively, in thex-y-plane, with all otherinput parameters remaining unchanged. Don t forget the drawing, next page!Copyrightc 2017 University of #2 Practice Problem II (a) =20P: Use vector addition of forces F1 exerted by Q1 and F2 exerted by Q2 to get net force vector F : F = F1 + F2 where F1 = $%&'%( * F2 = $(&'(( + r1 = | r1 | and 1 = (1/r1) r1, r2 = | r2 | and 2 = (1/r2) r2 Here, r1 is the vector drawn from charge Q1 (generating the electric force) to point charge q (subjected to the electric force).))))

5 Likewise, r2 is the vector drawn from charge Q2 (generating the electric force) to point charge q (subjected to the electric force). Step 1: [5P] Drawing for Part (a): Q1 [1P], Q2 [1P], F1 [1P], F2 [1P], F [1P] Force vectors need not be to scale, but must point in the correct direction: ( x). Step 2: See Drawing for Part (a) below! [2P] r1 = (xq, yq, zq) (x1, y1, z1) = (0, 0, 0)m ( , 0, 0)m = (+ , 0, 0) m [2P] r2 = (xq, yq, zq) (x2, y2, z2) = (0, 0, 0)m (+ , 0, 0)m = ( , 0, 0) m Step 3: [1P] r1 = | r1 | = [ + 02 + 02]1/2 m = m 1 = (1/r1) r1 = (1 ) (+ , 0, 0) = (+ 1, 0, 0) [1P] r2 = | r2 | = [( )2 + 02 + 02]1/2 m = m 2 = (1/r2) r2 = (1 ) ( , 0, 0) = ( 1, 0, 0) Step 4: [3P] F1 = $%&'%( *=9 102(45.))

6 + *789)(;*/*=) *78>( )( +1,0,0 N F1 = (F1x , F1y , F1z ) = ( , 0 , 0) N [3P] F2 = $(&'(( +=9 102(; *789)(;*/*=) *78>( )( 1,0,0 N F2 = (F2x , F2y , F2z ) = ( , 0 , 0) N Step 5: [2P] F = F1 + F2 = (F1x , F1y , F1z ) + (F2x , F2y , F2z ) = (( )+ ( ), 0+0, 0+0) N F = (Fx , Fy , Fz ) = ( , 0, 0) N Step 6: [1P] F = |F| = (Fx2 + Fy2 + Fz2 ) = | Fx| = Exam #2 Practice Problem II (a), cont d: Alternative Route: Step 1: [5P] same as above. Steps 2-4 (Alternative Route): [1P] r1 = distance(q, Q1)= |xq x1| = |0 ( )|m = [2P] |F1| = |$%&|'%( =9 102|(45.))))))

7 + *789)(;*/*=) *78>|( )( N = [2P] q is attracted by Q1 , since Q1 and q have opposite signs: Q1 <0 and q>0. F1 points along x-axis towards Q1, , in ( x)-direction [1P] F1 = (F1x , F1y , F1z ) with F1x = < 0 F1x = |F1| = and F1y = F1z = 0 [1P] r2 = distance(q, Q2) = |xq x2| = |0 (+ )|m = [2P] |F2| = |$(&|'(( =9 102|(; *789)(;*/*=) *78>|( )( N = [2P] q is repelled by Q2 , since Q2 and q have the same sign: Q2 >0 and q>0. F2 points along x-axis away from Q2, , in ( x)-direction [1P] F2 = (F2x , F2y , F2z ) with F2x < 0 F2x = |F2| = and F2y = F2z = 0 Step 5: [2P] same as above.))))

8 Step 6: [1P] same as above. Exam #2 Practice Problem II (b) = 15P: Use vector addition of electric fields E1, generated at P by Q1, and E2, generated at P by Q2, to get net electric field vector E at location P: E = E1 + E2 where E1 = $%'%( * E2 = $('(( + r1 = | r1 | and 1 = (1/r1) r1, r2 = | r2 | and 2 = (1/r2) r2 Here, r1 is the vector drawn from charge Q1 (generating the electric field) to observation point P. Likewise, r2 is the vector drawn from charge Q2 (generating the electric field) to observation point P.))))

9 Step 1: [ ] Drawing for Part (b): Q1 [1P], Q2 [1P], E1 [1P], E2 [1P], E [1P] E-field vectors need not be to scale, but must point in the correct direction: ( x). Step 2: See Drawing for Part (b) below! [2P] r1 = (xP, yP, zP) (x1, y1, z1) = ( , 0, 0)m ( , 0, 0)m = (+ , 0, 0) m [2P] r2 = (xP, yP, zP) (x2, y2, z2) = ( , 0, 0)m (+ , 0, 0)m = ( , 0, 0) m Step 3: [1P] r1 = | r1 | = [ + 02 + 02]1/2 m = m 1 = (1/r1) r1 = (1 ) (+ , 0, 0) = (+ 1, 0, 0) [1P] r2 = | r2 | = [( )2 + 02 + 02]1/2 m = m 2 = (1/r2) r2 = (1 ) ( , 0, 0) = ( 1, 0, 0) Step 4: [2P] E1 = $%'%( *=9 102(45.))

10 + *789)( )( +1,0,0 N/C E1 = (E1x , E1y , E1z ) = ( 405. , 0 , 0) 10?N/C [2P] E2 = $('(( +=9 102(; *789)( )( 1,0,0 N E2 = (E2x , E2y , E2z ) = ( 120. , 0 , 0) 10?N/C Exam #2 Practice Problem II (b), cont d: Step 5: [2P] E = E1 + E2 = (E1x , E1y , E1z ) + (E2x , E2y , E2z ) = (( 405.)+ ( 120.), 0+0, 0+0) 10?N/C E = (Ex , Ey , Ez ) = ( 525., 0, 0) 10?N/C Step 6: [ ] E = |E| = (Ex2 + Ey2 + Ez2 ) = | Ex| = 525. 10?N/C Alternative Route: Step 1: [ ] same as above. Steps 2-4 (Alternative Route): [1P] r1 = distance(P, Q1)= |xP x1| = |( ) ( )|m = [1P] |E1| = |$%|'%( =9 102|(45.))))))