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NCERT Class XII Chapter
Electric Charges and Fields
Questions With Solutions

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Question : 34
Total: 34
Suppose that the particle in question 33 is an electron projected with velocity vx = 2.0 × 106 m s–1. If E between the plates separated by 0.5 cm is 9.1 × 102 N C–1, where will the electron strike the upper plate?
(|e| = 1.6 × 10–19 c, me = 9.1 × 10–31 kg)
Solution:  
If the electron is released just near the negatively charged plate, then y = 0.5 cm and hence
y =
qEL2
2mvx2
 or L2 =
3mvx2y
qE
or L2 =
2×9.1×10−31‌Kg×(2×106ms−1)2×0.5×10−2m
1.6×10−19C×9.1×102NC−1

or L2 = 2.5 × 10−4 or L = 1.6 × 10−2 m = 1.6 cm
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