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Suppose that the particlein Question 1.33 is an electronprojected with velocity v_x =2.0 xx 10 ^(6) ms^(-1).IfE between the plates separated by 0.5 cm is 9.1 xx 10 ^(2) N//C,where will the electronstrike the upper plate ?(|e| =1.6 xx 10^(-19)C, m_e =9.1 xx 10 ^(-31) kg ) |
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Answer» Solution :For electron particle ` q=|e| =1.60 xx 10 ^(-19)C and ` mass of electron`m_e= 9.1 xx 10 ^(-31)kg. ` Makeover , it is given that `v_x =2.0 xx 10 ^(6)MS ^(-1)and E= 9.1 xx 10 ^(2)N//C ` Letafter covering a distance L the electron STRIKES the UPPER plate after suffering a DEFLECTION `y=0.5 cm = 5XX 10 ^(-3)m. `Then from the relation ` "" y= (qEL^(2))/( 2mv_x^(2)) , ` we have ` L =v_x xx sqrt((2my)/(qE) )=(2xx10^(6)) xx sqrt((2xx9.1 xx10^(-31)xx 5xx 10 ^(-3))/( 1.60 xx10 ^(-19)xx 9.1 xx 10 ^(2)))=1.58 xx 10 ^(-2)m = 1. 6 ` cm. |
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