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InterviewSolution
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An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude `2.0xx10^(4)N//C(Fig.a)` Calculate the time it takes to fall through this distance starting from rest. If the direction of the field is reversed (fig .b) keeping its magnitude unchanged , calculate the time taken by a proton to fall through this distance starting from rest. |
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Answer» The field is upward , so the negatively charged particle , electron experiences a downward froce of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is ae `=(eE)/(me)` Starting from rest , the time required by the electron to fall through a distance h `t_(e)=sqrt((2h)/(ae))=sqrt((2hme)/(eE))=sqrt((2xx1.5xx10^(-2)xx9.11xx10^(-31))/(1.6xx10^(-19)xx2.0xx10^(-4)))=2.9xx10^(-9)S`. The field is downward , and the positively charged proton experiences a downward force of magnitude eE . The acceleration of the proton is `aP=(eE)/(m_(p))` Hence , the time taken by the proton `t_(p)=sqrt((2h)/(a_(p)))=sqrt((2hm_(p))/(eE))=sqrt((2xx1.5xx10^(-2)xx1.67xx10^(-27))/(1.6xx10^(-19)xx2.0xx10^(-4)))=1.3xx10^(-7)sec`. |
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