Question

A 120 V battery is connected across two parallel metal plates of area 28.5 cm" and separation 7.20 mm. A beam of alpha particles (charge +2e, mass 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.50 kV and enters the region between the plates perpendicular to the electric field, as shown in (Figure 1).

Answer 1

00 am Paige Ans 0 Given -Pol- difference b/w plotes electric field b/w plity = 0 129 12 -7.20x10 0 E 16,66 810+ v/m ... Electric force on oder a particle Fe = VE 2e x 16.66x10^N noon 22x1,19 x 10-19 x 1666 x 1013 Fe 53.312 x10-16N Since the particle is acclerated through I.sour Potential difference, so the energy so the energy gained by a part -ide will be OG U = 1.50x10²x2x1.9816 U = 5.7 x 10 76 J This energy will be converted energy into Kinetic so my² - 5.7x10-16 2 V 5.7x40-16x2 m H.4 810-16 6.64*10-27 (²1.4410" =7V = 4.13 x 103 m/s

of magnetic force be B. 8.00 am Page ① 8.30 let the magnitude Magnetic force on a FB 9 VB 9.00 particle 9.30 10.00 10.30 should +2x1.9 x 10-19 for particle to pass undeyle be equal and Teel a list x 4.3 XLOS XB lected these opposite Hence forces 11.00 11.30 12.00 noon = 53.312x 10-16 3.8 X10 TX105X4.13 x B B 53.312 810161019x10-5 12.30 1.00 471383.8 1.30 B 3.262 x 10-21 2.00 2.30 (B) 3.00 3.30 of 4.00 The direction of force field should be shsuch that it applies gerce on x' particle in oppost dircetion electric field. The force can be cal right hand thums hele and can be found to be cerrect ansaver is • The magnetic field is directed out culated by 4.30 ou q page. 5.00 of page. 5.30 6.00 Evening