Question

(1 point) Two parallel plates are separated by a distance of 12.3 mm are connected to a 9 volt battery. A proton is emitted from positive plate with an initial speed of 1.62 x 10 m/s. (A) What is the speed of the proton when it is at a potential of 4.8 volts? (B) At what distance from the negative plate is the proton when it is at a potential of 4.8 volts? (C) What is the magitude of the electric field between the plates? (D) What is the electric potential of the proton when it is 8.41 mm from the negative plate? (E) How fast is the proton moving when it is 8.41 mm from the negative plate?

Answer 1

solution Given 9V et OV d= 12-3 mm v 9 Volts v=1:62810 m/s K 12.3mm YA x 0 blu - to AGO we know the electric field intensity between two oppesitely charged (but equal) plates is constant lie where q: charge on on the plates Area of the plates to: Permittivity of free space potential difference between the plates is given by A: V Εχ ماهمام x= 12.3 mm 9 E = how V= gu ( given) E x12:3 x 103 2 7.31 X10 we have a function v= f(x), (7.3) X102).X proton is at ve 4.2 Volt 7.31 X102.7 V = А اعمایی 4.8 = 4.8 2= x102 m. 7.31 25 4.8 -2 3 x 10 x 10 mm 7.31

à 6.57 mm given initial speed of proton 1.62 x 104 m/s how from third equation of motion e²= U= 2 u + 2 ax Acceleration: וד 2 a= qЕ m m -19 1.6810 X 7.31 X10 9.:1 X10-31 Y proton = 1.6X1619 9.109384 x 10-31 mproton 31 -17 x 10 x 10 a= (1687-31) a 1:28 x 10 x 1014 m/s2 <?= (1.62 x1042 + 2 x 1-28x1014X6.57xjö e² re? x1011 2.62 X 108 + 16.8 X10 2.62 X 109 + 16800 X 100 16 802.62 X108 22. Us 129.6 x1 of mis 1:296 X106 Ve mis B when proton is at v= 4.8, its distance from positive plate is x+= 6.57 mm, there-fore its distance from from negative plate is (12.3 -6.57 mm x-

2- = 5.73 mm ull we already calculated this and got (in the Introduction part) 9 V 12.3 x103 m Es v. E = 731.7. Volts/m all we formulated potential as v (n) = 731.7 x x 2 = 8.41 mm x103 V = 731.7 X 8.41 x103 Volts o at 8.41 XIO m V = 6.15 Volts wl 2 e2= + 2 as 14 m/s² from equation of motion. u² we have u= 1.62 x 10tmis e given) a = 1:28 X10 ( Part A s = 8.41 x103 m (given in Part E E) (1-62 x104)?- + 2 x 1.28 xiolt x 8.41 X10 2.62 X10 + 21:53 X1011 + 21530 X108 21532.62 x108 146.74 xlot mis v 1.4674 X106 mio !! 22 e? xioa e² 8 2.62 X 10 re 12 v = =