ent Previewer ollowing ligure. Assume that the held between the platles i form a

Question

ent Previewer ollowing ligure. Assume that the held between the platles i form and directed veitically downwand and that the S d-5.00 em hit one of the plates? and ne directio the irection-Select c) Compare the patths traveled by the electron and the proton and explain the dieences a) DscusE whay it is reasonable to ignoe the eftects of gravity for each p Perg charge@.--540at the ongn and poin chargeQ:..275rCisthe "wis..320cm.pwt P raxis at y #430cm. 0athe 7- i. 03/30/2016 09 ent Previewer http://www.webassign netvicgibberenjecal

Explanation / Answer

There is no horizontal force hence,

d2= v0x*t

t= d2/v0--------------------(1)

Along vertical,

h= v0y*t + 1/2at^2

d1/2 = 0*t +1/2*a*(d2/v0)^2

Plugging vales,

0.05/2 = 1/2*a*(0.025/(1.95*10^6))^2    => a= 3.04*10^14 m/s^2

Applying Newton’s second law vertically,

Fe - Fg = ma

eE – mg = ma

E = (ma+mg)/e = (9.1*10^-31*3.04*10^14 + 9.1*10^-31*9.8)/(1.6*10^-19) = 1729 N/C = 1.73*10^3 N/C

Check velocity I have taken.

Yes it will hit negative plate.

E1= E1y= kq1/y^2 = (9*10^9*5.40*10^-9)/(0.0320^2) = 4.75*10^4 N/C

E1x= 0 N/C

Thus E1= 0 N/C i + 4.75*10^4 N/C j

E2= kq1/y^2 = (9*10^9*2.75*10^-9)/(0.0320^2+ 0.0450^2) = 8.12*10^3 N/C

= tan^-1(4.50/3.20) = 54.6 deg

E2x= E2cos = (8.12*10^3)cos54.6 = 4.7*10^3 N/C

E2y= E2sin = (8.12*10^3)sin54.6 = 6.6*10^3 N/C

E2= 4.7*10^3 N/C i + 6.6*10^3 N/C j

Enetx = E1x + E2x = 0 + 4.7*10^3 N/C = 4.7*10^3 N/C

Enety= E1x + E2x = 4.75*10^4 N/C + 6.6*10^3 N/C = 5.4*10^4 N/C

Enet = sqrt[(4.7*10^3)^2 + (5.4*10^4)] = 4.7*10^3 N/C

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