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Suppose that the particlein Q.33 is an electron projectedwith velocity v_(x) = 2.0xx10^(6) ms^(-1). If E between the platesseparatedby 0.5 cm is 9.1xx10^(2) N//C, where will the electron strike the upperplate ? (|e| = 1.6xx10^(-19) C,m_(e) = 9.1xx10^(-31) kg). |
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Answer» Solution :Here,`v_(x) = 2.0xx10^(6) ms^(-1), E = 9.1xx10^(2) N//C,d = 0.5 cm = 5xx10^(-3)m` `q = e = 1.6xx10^(-19) C, m_(e) = 9.1xx10^(-31) kg` The electron will strikethe upperplate at its other endat `x = L` is SOON as itsdeflection `y = (d)/(2) = (5xx10^(-3))/(2) = 2.5xx10^(-3) m` From `y = (qE L^(2))/(2 mv_(x)^(2))` `L = sqrt((2my)/(qE)) xx v_(x) = sqrt((2xx9.1xx10^(-31)xx2.5xx10^(-3))/(1.6xx10^(-19)xx9.1xx10^(2))) xx2xx10^(6) = 1.12xx10^(-2) m = 1.12 cm` Hence the electronwill strikethe upperplateat itsother end, if length of plate is 1.12 cm. |
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