Two 2.5 cm -diameter disks face each other, 2.0mm apart. They are charged to plu

Question

Two 2.5 cm -diameter disks face each other, 2.0mm apart. They are charged to plus or minus 14 nC A)What is the electric field strength between the disks? B)A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk? I have tried doing this problem over and over again I cant seem to get the right answer i have been doing part A) E= 14*10^-9/ (0.01m)^2 (8.85*10^-12)(pi) and this isnt working ;( please explain what im doing wrong WILL RATE

Explanation / Answer

a) E=Q/(e0*A), where A=(pD^2)/4 ==> E=4Q/ (pe0D^2)=4*14e-9/(p*8.854e-12*0.25^2) E=32242 V/m b) e=1.6e-19 proton charge, m=1.6726e-27 proton mass, d=0.02m Ekin(neg disc)=Epot(pos disc) 0.5mv^2=eU, where U=dE ==> v=v(2edE/m)= v(2*1.6e-19*0.02*32242/1.6726e-27)= 351.240 km/s

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