Two insulating plates are both uniformly charged in such a way that the potential difference between them is V2 -V1 = 20 V. (i.e. plate 2 is at a higher potential). The plates are separated by d = 0.1 m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e = 1.6 × 10-19 C, me = 9.11 × 10-31 kg) ?
sudhanshu , 11 Years ago
Grade 12
2 Answers
SHAIK AASIF AHAMED
Last Activity: 10 Years ago
Hello student,
1/2 mv2= eV v=sqrt(2eV/m) = = 2.65 × 106m/s
Thanks and Regards
Shaik Aasif
askIITians faculty
Janahvi
Last Activity: 5 Years ago
e(V2-V1)=1/2 mv*2
1.6×10*-19 ×20 = 1/2 9.11×10*-31×v2
V*2= 2×32×10*-19/9.11×10*-31
V*2=64×10*-19/9.11×10*-31
V=√7.02×10*12 now applying square root
You get v= 2.65 ×10*6
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