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NCERT Class XII Chapter
Electric Charges and Fields
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Question : 34 of 34
Marks: +1, -0
Suppose that the particle in question 33 is an electron projected with velocity vxv_x = 2.0 × 10610^6 m s−1s^{-1}. If E between the plates separated by 0.5 cm is 9.1 × 10210^2 N C−1C^{-1}, where will the electron strike the upper plate?
(|e| = 1.6 × 10−1910^{-19} c, mem_e = 9.1 × 10−3110^{-31} kg)
Solution:  
If the electron is released just near the negatively charged plate, then y = 0.5 cm and hence
y = qEL22mvx2\frac{qEL^2}{2mv_x^2} or L2L^2 = 3mvx2yqE\frac{3mv_x^2 y}{qE}
or L2L^2 =
2×9.1×10−31Kg×(2×106ms−1)2×0.5×10−2m1.6×10−19C×9.1×102NC−1\frac{2 \times 9.1 \times 10^{-31} \mathrm{Kg} \times (2 \times 10^6 \mathrm{ms}^{-1})^2 \times 0.5 \times 10^{-2} \mathrm{m}}{1.6 \times 10^{-19} \mathrm{C} \times 9.1 \times 10^2 \mathrm{NC}^{-1}}
or L2L^2 = 2.5 × 10−410^{-4} or L = 1.6 × 10−210^{-2} m = 1.6 cm
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