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Suppose that the particle in Exercise in 1.33 is an electron projected with velocity v_x = 2.0 xx 10^6 ms^(-1) . If E between the plates separated by 0.5 cm is 9.1 xx 10^2 N/C, where will the electron strike the upper plate ? (|e| = 1.6 xx 10^(-19) C, m_(e) = 9.1 xx 10^(-31) kg) |
Answer» Solution :Here, electron strikes UPPER plate and so upper plate must be positive.  Now from equation (5) of above example no. 33 we have, `y=(QEX^(2))/(2mv_(s)^(2))` `therefore x^(2) = (2ymv_(x)^(2))/(qE)` `therefore x=((2ymv_(x)^(2))/(qE))^(1//2)` `therefore x = 1.125 XX 10^(-2)` m |
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