Fundamentals of Physics chapter 21 solutions

Chegg costs money, GradeSaver solutions are free!

Chapter 21 - Coulomb's Law - Problems - Page 625: 4

Answer

Work Step by Step

As we know that current i is given as $i=\frac{q}{t}$ where q is charge and t is time or $q=it$ putting the values $q=(2.5\times10^4)(20\times10^{-6})$ $q=0.50 C$

Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.

Halliday, Resnick, and WalkerFundamentals of Physics 10eProblem AnswersVolume 2Chapter 21 Answers10.50020.37531.39 m40.50 C52.81 N6(a) 4.910-7kg;(b) 7.110-11C7-4.0080.3759(a) -1.00C;(b) 3.00C10(a) -2.83;(b) no11(a) 0.17 N;(b)0.046 N12(a)83C;(b) 55C13(a) -14 cm;(b) 014(a) 9.0;(b)2515(a) 35 N;(b) -10;(c) -8.4 cm;(d) +2.7 cm16(a) positive;(b) +9.017(a) 1.60 N;(b) 2.77 N181.33319(a) 3.00 cm;(b) 0;(c) -0.44420(a)4;(b) +16213.810-8C22(a) 1.92 cm;(b) less than

23(a) 0;(b) 12 cm;(c) 0;(d) 4.910-26N24(a) 8.9910-19N;(b) 625256.31011262.8910-9N27(a) 3.210-19C;(b) 2282.25102029(a) -6.05 cm;(b) 6.05 cm30(a) 2.00 cm;(b) 9.2110-24N31122 mA32+13e331.3107C34(a) 0.654 rad;(b) 0.889 rad;(c) 0.988 rad35(a) 0;(b) 1.910-9N36(a) positron;(b) electron37(a)9B;(b)13N;(c)12C38+16e391.3110-22N40-2.2541(a) 5.71013C;(b) cancels out;(c) 6.0105kg42(b) 2.410-8C43(b) 3.1 cm4411.9 cm450.19 MC46(a) (3.5210-25N)i^;(b) 047-45C48(a) 3.60N;(b) 2.70N;(c) 3.60N

493.8 N50(a) (L/2)(1 +kqQ/Wh2);(b) (3kqQ/W)0.551(a) 2.001010electrons;(b) 1.331010electrons5211.1C53(a) 8.99109N;(b) 8.99 kN549.0 kN55(a) 0.5;(b) 0.15;(c) 0.8556(a) 1.251013electrons;(b) from you to faucet;(c) positive;(d) from faucet to the cat;(e) stroking the cat transfers electrons from you to thefur, which then induces charge in the cat’s body, withnegative charge on the surface away from the strokedregion; if you bring your positive hand near thenegative nose, electrons can spark across the gap571.7108N58(a) (89.9 N)i^;(b) (-2.50 N)i^;(c) 68.3 cm;(d) 059-1.321013C60061(a) (0.829 N)i^;(b) (-0.621 N)j^62(a) 6.1610-24N;(b) 208632.210-6kg641.210-5C654.6810-19N66-5.1 m67(a) 2.72L;(b) 0681018N69(a) 5.1102N;(b) 7.71028m/s2700.70771(a) 0; (b) 3.43109m/s2721.6 nm

73(a) 2.19106m/s; (b) 1.09106m/s; (c) decrease741.3 days754.161042Chapter 22 Answers1---2(a) 6.410-18N;(b) 20 N/C3(a) 3.071021N/C(b) outward4(-6.39105N/C)i^556 pC60.111 nC7(1.02105N/C)j^809(a) 1.3810-10N/C;(b) 18010(a) 34 cm;(b) 2.210-8N/C11-30 cm12(a) 3.9310-6N/C;(b) -76.413(a) 3.6010-6N/C;(b) 2.5510-6N/C;(c) 3.6010-4N/C;(d) 7.0910-7N/C;(e) As the proton nears the disk, the forces on it fromelectrons esmore nearly cancel.14(a) 2.72L15(a) 160 N/C;(b) 4516(a) 67.8;(b) -67.817(a) -90;(b) +2.0C;(c)1.6C18qd3/40z519(a)qd/40r3;(b)90200.9821---

22(a)1.7210-15C/m;(b)3.8210-14C/m2;(c)9.5610-15C/m2;(d)1.4310-12C/m3230.50624(a) 0;(b) 0;(c) 0.707R;(d) 3.46107N/C25(a) 1.62106N/C;(b) -4526(a) 20.6 N/C;(b)9027(a) 23.8N/C;(b)90281.70 cm291.5730-4.19Q31(a)5.1910-14C/m;(b) 1.5710-3N/C;(c)180;(d) 1.5210-8N/C;(e) 1.5210-8N/C32(a) 12.4 N/C;(b) 9033---346.3103N/C350.346 m362.410-16C3728%386.9 cm39-5e40(a) 7.12 cm;(b) 28.5 ns;(c) 0.11241(a) 1.5103N/C;(b) 2.410-16N;(c) up;(d) 1.610-26N;(e) 1.5101042(a) 4.810-13N;(b) 4.810-13N433.511015m/s2

44(a) 2.0310-7N/C;(b) up456.610-15N46(a) 1.0210-2N/C;(b) west47(a) 1.921012m/s2;(b) 1.96105m/s48(a) 1.161016m/s2;(b) 3.941016m/s2;(c) 3.971016m/s2;(d) because the net force due to the charged particlesnear the edge of the disk decreases49(a) 0.245 N;(b) -11.3;(c) 108 m;(d) -21.6 m50(a) (-2.11013m/s2)j^;(b) (1.5105m/s)i^- (2.8106m/s)j^51(a) 2.610-10N; (b) 3.110-8N; (c) moves to stigma52

Upload your study docs or become a

Course Hero member to access this document

Upload your study docs or become a

Course Hero member to access this document