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Question Number 55431 by necx1 last updated on 24/Feb/19
2 plane parallel conducting plates are  held horizontal,one above the other in  a vacuum.Electrons having a speed of  6×10^6 m/s and moves normally to the  plates enter the region between them  through a hole in the lower plate which  is earthed.What potential must be  applied to the other plate so that the  electrons just fails to reach it?What  is the subsequent motion of these  electrons?  (ratio of charge to mass of electron=  1.8×10^(11) Ckg^(−1) )
$$\mathrm{2}\:{plane}\:{parallel}\:{conducting}\:{plates}\:{are} \\ $$$${held}\:{horizontal},{one}\:{above}\:{the}\:{other}\:{in} \\ $$$${a}\:{vacuum}.{Electrons}\:{having}\:{a}\:{speed}\:{of} \\ $$$$\mathrm{6}×\mathrm{10}^{\mathrm{6}} {m}/{s}\:{and}\:{moves}\:{normally}\:{to}\:{the} \\ $$$${plates}\:{enter}\:{the}\:{region}\:{between}\:{them} \\ $$$${through}\:{a}\:{hole}\:{in}\:{the}\:{lower}\:{plate}\:{which} \\ $$$${is}\:{earthed}.{What}\:{potential}\:{must}\:{be} \\ $$$${applied}\:{to}\:{the}\:{other}\:{plate}\:{so}\:{that}\:{the} \\ $$$${electrons}\:{just}\:{fails}\:{to}\:{reach}\:{it}?{What} \\ $$$${is}\:{the}\:{subsequent}\:{motion}\:{of}\:{these} \\ $$$${electrons}? \\ $$$$\left({ratio}\:{of}\:{charge}\:{to}\:{mass}\:{of}\:{electron}=\right. \\ $$$$\left.\mathrm{1}.\mathrm{8}×\mathrm{10}^{\mathrm{11}} {Ckg}^{−\mathrm{1}} \right) \\ $$
Commented by necx1 last updated on 24/Feb/19
urgently needed please.
$${urgently}\:{needed}\:{please}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Feb/19
potential lower plate =0  potential upper plate=v  eEd=(1/2)mv_(velocity) ^2   E=((v−0)/d)  ev=(1/2)mv_(vel) ^2   v=(1/2)×(1/(((e/m))))×v_(vel) ^2   v=(1/2)×(1/(1.8×10^(11) ))×36×10^(12)   v=((10)/(36))×((36×10)/1)=100 volt  pls check
$${potential}\:{lower}\:{plate}\:=\mathrm{0} \\ $$$${potential}\:{upper}\:{plate}={v} \\ $$$${eEd}=\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{velocity}} ^{\mathrm{2}} \\ $$$${E}=\frac{{v}−\mathrm{0}}{{d}} \\ $$$${ev}=\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{vel}} ^{\mathrm{2}} \\ $$$${v}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\left(\frac{{e}}{{m}}\right)}×{v}_{{vel}} ^{\mathrm{2}} \\ $$$${v}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{1}.\mathrm{8}×\mathrm{10}^{\mathrm{11}} }×\mathrm{36}×\mathrm{10}^{\mathrm{12}} \\ $$$${v}=\frac{\mathrm{10}}{\mathrm{36}}×\frac{\mathrm{36}×\mathrm{10}}{\mathrm{1}}=\mathrm{100}\:{volt} \\ $$$${pls}\:{check} \\ $$$$ \\ $$
Commented by necx1 last updated on 24/Feb/19
very correct. Thank you sir.
$${very}\:{correct}.\:{Thank}\:{you}\:{sir}. \\ $$

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