What is the electric field strength between the disks?

Two 4.0cm-diameter disks face each other, 1.5mm apart. They are charged to +/- 11nC.

Answer 1

a) E=Q/(ε₀*A), where A=(πD^2)/4 ==>
E=4Q/ (πε₀D^2)=4*11e-9/(π*8.854e-12*0.04^2)
E=988652 V/m

b) e=1.6e-19 proton charge, m=1.6726e-27 proton mass, d=0.0015 m
Ekin(neg disc)=Epot(pos disc)
0.5mv^2=eU, where U=dE ==>
v=√(2edE/m)=
√(2*1.6e-19*0.0015*988652/1.6726e-27)=
532.655 km/s

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