Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 1 1. Particles A and B are electrically neutral and are separated by 5.0 μm. If 5.0 x 10 6 electrns are transferred frm particle A t particle B, the magnitude f the electric frce between them is: A) 2.3 x 10-4 N B) 1.3 x 10-4 N C) 1.0 x 10-4 N D) 1.0 x 10-3 N E) 5.0 x 10-3 N F = kkqq2 rr 2 = 9 109 ( 5 10 6 1.6 10 19 ) (5 10 6 ) 2 = 2.3 10 4 N Sec# Electric Charge - Culmb's Law 2. Tw charges, +2.00 nc and 2.00 nc, are psitined at the crdinates (1.00 m, 0) and (0, 1.00 m) respectively in an islated space. The magnitude and directin f the electric field at the rigin are: A) 25.5 N/C, 135 0 frm psitive x-axis B) 25.5 N/C, 225 0 frm psitive x-axis C) 36.0 N/C, 135 0 frm psitive x-axis D) 18.0 N/C, 45 0 frm psitive x-axis E) 5.25 N/C, 225 0 frm psitive x-axis E nnnnnn E 1 2μμc E 2 θθ 2μμc E = 9 109 2 10 6 1 2 ii + 9 109 2 10 9 1 2 ĵ = 18ii + 18ĵ E = 18 2 + 18 2 = 25.5 N/CC θ = tan 1 18 = 45 = 135 frm psitive x-axis 18 Sec# Electric fields - The Electric Field Due t a Pint Charge 3. An electric diple mment f magnitude p = 3.02 x 10-25 C.m makes an angle 64 with a unifrm electric field f magnitude E = 46.0 N/C. The wrk required t turn the electric diple by 90 is:
Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 2 A) 1.86 x 10-23 J B) 3.00 x 10-23 J C) 2.22 x 10-23 J D) 1.86 x 10-22 J E) 2.22 x 10-22 J WW aa = u = u f u i = EE p cs 154 ( EE p cs 64 ) = EE p (cs 64 cs 154 ) = 46 3.02 10 25 (0.438 + 0.898) =1.86 10-23 J Sec# Electric fields - A Diple in an Electric Field 4. What is NOT TRUE abut electric field lines: A) They are extended away frm a negative pint charge. B) They are the path fllwed by a unit psitive charge. C) Tw electric field lines never crss each ther. D) They are imaginary lines used t visualize the patterns in the electric field. E) The directin f the electric field at a pint in the field line is given by the tangent at that pint. Ans. A and B. Sec# Electric fields - Electric Field Lines 5. Figure 1 shws a Gaussian cube in regin where the electric field E is alng the y- axis. E = 30.0 ˆj N/C n the tp face and +20.0 ĵ N/C n the bttm face f the cube. Determine the net charge cntained within the cube. Fig# 1 y x z x = 1 m x = 3 m
Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 3 A) 1.77 10-9 C B) +1.77 10-9 C C) 3.98 10-9 C D) +3.98 10-9 C E) 3.54 10-10 C = E A = q end εε 0 = 30 4 20 4 = q end εε 0 q eeeeee = 200 8.85 10 12 = 1.77 10 12 C Sec# Gauss's law - Gauss's Law 6. Figure 2 shws shrt sectins f tw very lng parallel lines f charge, fixed in place and separated by L = 10 cm. The unifrm linear charge densities are +6.0 µc/m fr Line 1 and 2.0 µc/m fr Line 2. Find the x-crdinate f the pint alng the x-axis at which the net electric field due t the tw line charges is zer. Fig #2 Line 1 y Line 2 xx EE 2 E 1 x L/2 L/2 λ 1 = 6 μc 2μc = λ 2 A) +10 cm B) 10 cm C) 15 cm D) +15 cm E) Zer E 2 = EE 1 2kk λλ 1 rr 1 = 2kk λλ 2 rr 2
Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 4 λλ 1 LL 2 + x = λλ 2 x LL 2 6 x LL 2 = 2 LL 2 + x 4L = 4x x = L = 10 cm Sec# Gauss's law - Applying Gauss's Law: Cylindrical Symmetry 7. Figure 3 shws the magnitude f the electric field due t a sphere with a psitive charge distributed unifrmly thrughut its vlume. What is the value f the charge n the sphere? Fig# 3 A) 2 µc B) 1 µc C) 3 µc D) 4 µc E) 5 µc EE = k R 2 5 10 7 = 9 109 (0.02) 2 = 2.2 10 6 C Sec# Gauss's law - Applying Gauss's Law: Spherical Symmetry 8. Which f the fllwing statements is NOT TRUE? A) If a charged particle is placed inside a spherical shell f unifrm charge density, there is a nn-zer electrstatic frce n the particle frm the shell. B) The electric field is zer everywhere inside a cnductr in electrstatic equilibrium.
Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 5 Ans. A. C) Any excess charge placed n an islated cnductr in electrstatic equilibrium resides n its surface. D) The electric field just utside a charged cnductr in electrstatic equilibrium is perpendicular t its surface. E) The net electric flux thrugh any clsed surface is prprtinal t the net charge inside the surface. Sec# Gauss's law - A Charged Islated Cnductr 9. An electrn is released frm rest in a dwnward electric field f magnitude 1500 N/C. Determine the change in electric ptential energy f the electrn when it has mved a vertical distance f 20 m in this electric field. A) 4.8 10 15 J B) + 4.8 10 15 J C) 2.4 10 15 J D) + 2.4 10 15 J E) + 1.2 10 15 J UU = 9 VV = 9 ( Edd cs 180 ) = 9 E dd = ( 1.6 10 19 ) 1500 20 = 4.8 10 15 J Sec# Electric Ptential - Electric Ptential Energy 10. Figure 4 shws the x cmpnent f the electric field as a functin f x in a regin. The y and z cmpnents f the electric field are zer in this regin. If the electric ptential at x = 0 is 10 V, what is the electric ptential at x = 3 m? Fig. # 4 A) +40 V
Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 6 B) +30 V C) 30 V D) 40 V E) 10 V 3 VV = EE ddss = EE x ddx 0 = Area under the curve VV = 1 2 3 ( 20) = 30 V V f = 30 V+ V i = 40 V Sec# Electric Ptential - Calculating the Ptential frm the Field 11. A particle f charge 2 10-3 C is placed in an xy plane where the electric ptential depends n x and y as shwn in Figure 5. The ptential des nt depend n z. What is the electric frce (in N) n the particle? Fig# 5 A) + 5i ˆ 2j ˆ B) 5i ˆ+ 2j ˆ C) + 2i ˆ 5j ˆ D) 2i ˆ+ 5j ˆ E) + 5i ˆ+ 2j ˆ
Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 7 FF = 9 EE = 9 VV x x ii VV yy YY jj = 2 10 3 1000 0.4 = 5 ii 2jj NN ii 500 0.5 jj Sec# Electric Ptential - calculating the Field frm the Ptential 12. Hw much wrk is required t set up the arrangement f charges shwn in Figure 6, if = 3 µc? Assume that the charged particles are initially infinitely far apart and at rest. Take the ptential t be zer at infinity. Fig # 6 3 m 4 m A) +0.06 J B) 0.06 J C) +0.03 J D) 0.03 J E) 0.02 J WW aa = uu = uu ff uu ii = k 2 1 r 12 + 1 r 13 + 1 r 23 = 9 10 9 ( 3 10 6 ) 2 1 5 + 1 3 + 1 4 = 0.06 J Sec# Electric Ptential - Electric Ptential Energy f a System f Pint Charges
Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 8 13. Figure 7 shws tw different arrangements f 12 electrns fixed in space. In the arrangement (a) they are unifrmly spaced arund a circle f radius R and in (b) they are nn-unifrmly spaced alng an arc f the riginal circle. Which f the fllwing statements is TRUE? Fig# 7 A) The electric ptential at the center C is the same in bth (a) and (b). B) The electric ptential at the center C is larger in (b) than that in (a). C) The magnitude f electric field at the center C is the same in bth (a) and (b). D) The magnitude f electric field at the center C is smaller in (b) than that in (a). E) The electric ptential and the electric field at the center C are bth zer in (a). Ans. A Sec# Electric Ptential - Ptential Due t a Grup f Pint Charges 14. A parallel-plate capacitr has a plate area f 0.20 m 2 and a plate separatin f 0.10 mm. T btain an electric field f 2.0 x 10 6 N/C between the plates, the magnitude f the charge n each plate shuld be: A) 3.5 x 10-6 C B) 1.8 x 10-6 C C) 7.1 x 10-6 C D) 8.9 x 10-6 C E) 1.4 x 10-6 C = CCCC = εε 0AA dd EEEE = 8.85 10 12 0.2 2 10 6
Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 9 = 3.54 10 6 C Sec# Capacitance - Calculating the Capacitance 15. Figure 8 shws an arrangement f fur capacitrs, 500 µf each with air between the plates. When the space between the plates f each capacitr is filled with a material f dielectric cnstant 2.50, the vltmeter reads 1000 V. The magnitude f the charge, in Culmbs, n each capacitr plate is: Fig#8 A) 1.25 B) 0.250 C) 20.5 D) 52.5 E) 100 V CC 1 CC 1 CC 1 = kc = 2.5 500 10 6 = 1.25 10 3 F CC 1 CC 1 1 = C 1 V = 1.25 10 3 10 3 = 1.25 C Sec# Capacitance - Capacitrs in Parallel and in Series 16. Tw parallel-plate capacitrs with the same capacitance but different plate separatin are cnnected in series t a battery. Bth capacitrs are filled with air. The quantity that is NOT the same fr bth capacitrs when they are fully charged is: A) The electric field between the plates B) The stred energy C) The ptential difference D) The charge n the psitive plate E) The dielectric cnstant Ans. A. Sec# Capacitance - Capacitrs in Parallel and in Series
Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 10 17. An air filled parallel-plate capacitr has a capacitance f 3.0 pf. The plate separatin is then increased three times and a dielectric material is inserted, cmpletely filling the space between the plates. As a result, the capacitance becmes 5.0 pf. The dielectric cnstant f the material is: A) 5.0 B) 10 C) 2.0 D) 4.0 E) 3.0 C 0 =3pF C 0 = ε 0A d C 1 = ε 0A = C 0 = 1pF 3d 3 C 2 = kkkk = 5 1pF = 5pF Sec# Capacitance - Capacitr with a Dielectric 18. A hallw spherical cnductr f inner radius 2.0 cm and uter radius 4.0 cm has a net charge f 4.0 nc. If a pint charge f 5.0 nc is placed at the center f the hallw spherical cnductr, the charge density n the uter surface f the cnductr is: A) + 50 nc/m 2 B) + 80 nc/m 2 C) 50 nc/m 2 D) 80 nc/m 2 E) + 40 nc/m 2 5nc in =5nc σ ut = ut A = 1 10 9 4ππ(0.04) 2 σ ut = 49.7 10 9 C 50 nc -4nc ut =1nc Sec# Gauss's law - Applying Gauss's Law: Spherical Symmetry 19. Tw small charged bjects attract each ther with a frce F when separated by a distance d. If the charge n each bject is reduced t ne-furth f its riginal value and the distance between them is reduced t d/2 the frce becmes:
Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 11 A) F/4 B) F/2 C) F/16 D) F E) F/8 F = kq 1q 2 d 2 F = k qq 1 qq 2 4 4 dd 2 = 2 k q 1 q 2 16 dd2 4 = k q 1 q 2 4dd 2 = F 4 Sec# Electric Charge - Culmb's Law 20. Eight identical spherical raindrps are each at a ptential V (assuming the ptential t be zer at infinity). They cmbine t make ne spherical raindrp whse ptential is: A) 4V B) 2V C) V/2 D) V/4 E) V/8 Vlume 2 = 8 vlume 1 4 3 ππr 2 3 = 8 4 3 ππr 1 3 R 2 = 2R 1 VV 2 = k 8 R 2 = k 8 2R 1 = 4k R 1 = 4 V Sec# Electric Ptential - Ptential Due t a Grup f Pint Charges
y y Line 1 Line 2 x x z x = 1 m x = 3 m L/2 L/2 Figure 1 Figure 2 Figure 3 Figure 4 Figure 5 3 m 4 m Figure 6 Figure 7 Figure 8
Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 1 Physics 102 Frmula sheet fr Secnd Majr iˆ, jˆ and kˆ are unit vectrs alng the psitive directins f x-axis, y-axis and z-axis respectively. kq1q 2 F =, F = q 2 0 E r v = v + at kq 1 2 Φ = E. da, E = 2 x x = v t + at r Surface 2 2 2 k 2kλ v = v + 2a(x x ) E = r, E = 3 R r qin σ σ φc = E.dA = ; E = ; E = ε0 2ε ε ε 0 = 8.85 10-12 C 2 /N.m 2 k k = 9.0 10 V =, W = - 9 N.m 2 /C 2 U e = -1.6 10 r -19 C B U V = V B - V A = - E.ds m e = 9.11 10-31 kg = m p = 1.67 10-27 kg q A 0 1 ev = 1.6 10-19 J V V V =, =, = g = 9.8 m/s 2 E x U = x kq q r 1 2 12 E y C =, C V 1 2 U = CV, 2 τ=p E y E z z ε 0A ab =, C = 4πε d b a, 1 2 u = 2 ε E, C = κ C 0,, U = p.e micr ( µ ) = 10-6 nan (n) = 10-9 pic (p) = 10-12