Plz solve 21,22,23,24,25,26

Plz solve 21,22,23,24,25,26 Calculate the magnitude and direction of the electric fveid. which keeps a proton just floating. Given that mass Of proton 1.67 x 10-27 kg. charge on proton 1.6 [Ans. 1-023 x 10¯' N (vetticaity up ward)) c and g = 9.8 m s-2 An electron above the earth is balance by the gravitational force and the eiectre [Ans. x NC-n field of the earth. An electron is released with a velocity of 5 x 106 m in an electric Of N which has been applied So as to oppose its motion. What distance would the electron travel and how much time could it take before it is brought to rest? (Ans. 7.11 x 10-2 m. 2.844 x 21. An electron falls through a distance of 1 _5 cm in a uniform electric field of magnitude 2.0 x 104 N [Fig.(a)) The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance Compute the time of fall in each case. Given that mass of electron 9.1 x 10-31 kg and mass of = 1.67 x 10-2' kg. [Ans. (a) 2.92 x (b)1.25 x 22. A charged particle of mass 1 g is suspended through 104 N C—l. If the particle thread of length 40 cm in a horizontal electric field stays at a d istance of 24 cm from the wall in leiun-i, find the charge on the (Ans. x to-7 panicle. are placed 18 cm apart. Find the positim 23. Two point charges of + 20 BC and + of the point, where electric field is 20 gCcharges (between the two charges)) [Ans. At a distance O. corners of a square of side 2 24. Fig. Shows four point cha •on of the electric field at the cm. Find the magnitude an 0.02 gc. centre O of the square 9 Nm2C—2 Use — Ans. 9v'Z x 10s N C -2 , (parallel to side BA)I equal to q, are placed at the three corners of a square of side 25. Three cha ric field at the fourth corner of the square. (Ans. — a. Find h •g -angled triangle, the right angle being at point B. Charges of 26. AB 2 11.52, +3.24 ggC are placed at points A, B and C If AB=4cm C = 3 cm, calculate the magnitude and direction of the resultant electric field at foot D of the perpendicular drawn from point B on the side AC. [Ans. 180v'Z N C -1, inclined equally with both DA and BD (when produced)) ON ANELECTRCOIPOLE 27K' Two point charges of 2 VIC but opposite in sign are placed 10 cm apatt. Calculate the electric field at a point distant 10 cm ftom the mid-point on the axial hoe of the dipole, (Ans. 6.4 x 106 NC-I) p AGE

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Q23 two charges 20μC and 80μC are placed at 18cmThe electric field will be zero atK×20x2=K×80(18-x)218-xx=2x=6cm=0.06m (from charge 20micro C)REGARDS
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