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An electron falls through distance of 1.5 cm in a uniforth electric field of magnitude 2.0xx 10 ^(4) NC^(-1)The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance compute the time of falls in each case . Contrast the situation with that of 'free fall under gravity' |
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Answer» Solution :In the field is upwards , so the negativelycharged electron experiences a downward force of magnitude EE is the magnitude of the electric field. The acceleration of the electron is `a_e (aE)/( m_e) ` ,where `m_e` is the mass of the electron. Starting from rest , the time required by the electron to the fall through a distance h is given by `t_e =sqrt((2h)/( a_e) ) =sqrt((2hm_e)/(eE) ) ` For ` e=1.6 xx10^(-19) C,m_e =9.11 xx10 ^(-11) kg. ` `E=2.0xx10^(4)NC^(-1) ,h=1.5 x10^(-2) m,` `t_0 =2.9 xx10^(-9)s ` the field is downward and the positively CHARGED proton experiences a downwards force of magnitude eE. The acceleration of the proton is `a_p =(eE)/( m_p) , `where `m_p` is the mass of the proton `m_p =1.67 xx10 ^(37) `kg. The time of fall for the proton is `t_p =sqrt((2h)/(a_p)) =sqrt((2hm_p)/(eE) ` Thus ,the heavier particle (proton) takes a greater time to fall through the same distance . This is in basic contrast to the situation of free fall gravity where the time of fall is independant of the same mass of the body. Note that in this example we have ignored the acceleration due to gravity in a calculating time of fall . To see if this is justified let us calculate the acceleration of the proton in the given electric field. ` a_p =(eE)/( m_p) =(1.6xx10^(19))xx(2.0xx10^(4))/(1.67xx 10^(-37)) =1.9xx 10^(12) ms^(-2) ` Which is enormous compared to the value of `g(9.8 ms^(-2)) ` the acceleration due to gravity. the accelration of the electron is given greater. Thus the effect of acceleration due to gravity can be ignored in this example. |
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