Question

Chapter 19, Problem 16 GO The drawing shows a square, each side of which has a length of L = 0.250 m. Two different positive charges 41 and 42 are fixed at the corners of the square. Find the electric potential energy of a third charge 43 = -5.00 x 10-°C placed at corner A and then at corner B. B 91 = +1.50 x 10-C 92 = +4.00 x 10°C EPEA = EPEB

Answer 1

19) The electric potential energy of q3 at A and B is the sum of potential energy due to q1 and q2.Potential energy at A and B differs due to difference in charges and distance (r).

chapter 19 ; Problem 16 1st case B when charge 23 = -5x10*9 is placed at A. 93 A 0.250 m 22 = + 4 x 10-96 9,2+1-50X10-9 Diagonal of square = V2 x 0.250 m = 0.35m 9,92 potential energy of at az due to a, Vis - 1/aneo 4t Co q, az حب 0.35 1.50X10-9 x -5x10 9 lanco * 0.35 -21.428 *10-18 1 x 4R6o -210 4288 10-18 AT Eo

U₂ 3 = "laneo 9293 7 llaneox 4 X 10-9 x (-5) x 10-9 0.250 (-80 x 10-18) 42 Co U13 + U23 Total potential energy at (A) 'Vaneo (6-21-426)x 10-48 +(-80)x10-" = 9x109 6101428) 10-18 - -912.852 x 10-9 J at (B) 7 when the charge is 0.250m 93 B 0.35n 92.

4 43 42 43 + lanco ( Ві Фо. 1.5 х 10-4 х (-5) Хto-9 + MATEO 4x 10:4x65x10 о• 35 О• або б• або "Аке ( - 415) к но" + (-20 x 108 Як о* ( ( әр + -5442)од л - 784 • 18 хто- Ј

18) The work done is the change in potential energy. Here it is the difference between potential energy at B and A.

  

Chapter 19 Problem 18. Let Q=+460 4C 0.83 D q = +26540 If we move q foom 0.585 k A - B B work done change in potential energy That is, work done = PoE (B) - p.E(A). PoE (A) = AREO Q2 = '/anco Q Q 0.585 AE = 9x109 ( AGOX 10-6 * 2.5x 10-6 ) 0.585 11 9x109 ( 1,150 x 10-12 ) 0.585. - 3 17,692.307 X10

P.E at (B) 'larko Chele ) = '/axeo 460 X10-6 X 205x10-6 J 0.585 17 15 same as of at (A) Therefore, por at (B) = 17,692.307 x 10-35 work done = PIE (B) - P.E(A) as por (B) = PoE (A) wook done

19)The potential becomes 0 at the point.

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Chapter 19: Pooblem 22 The diagoam based on the question is A accooding to pythagores theoton ( 12 + 2². L = taq Am consider his on the dashed line at A. * electoo Static potential at A due to a is Potentiat (79) = "laxeo (-2) * potential (+29) (12 +22) "lanco (or when the total potential is zero. 291 4REO 1 +42 AREO () lanco [ 29 1+2² ak + 22 1+22

>> 22 +1 L² 2 242-2² = 1 L2=1 人二士|| L is im a bore Which means that below the negative change on the dashed line where total potential is zero.