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Consider an electron travelling with a speed V_(0) and entering into a uniform electric field vecE which is perpendicular to vecV_(0) as shown in the Figure. Ignoring gravity, obtain the electron 's acceleration velocity and position as functions of time. |
Answer» SOLUTION :  Speed of an electron `= V_(0)` Uniform electric FIELD = `VECE` (a) Electron.s acceleration : Force on electron due to uniform electric F = Ee Downward acceleration of electron due to electric field a =` (F)/(m) = -(eE)/(M)` Vector form `veca= -(eE)/(M)HATJ` (b) Electron.s velocity : Speed of electron in horizontal direction u = `V_(0)` From the equation of motion V= `mu` +at `V= V_(0)-(eE)/(M)t` vector form `vecV=V_(0)hatj- (eE)/(M)t hatj` (C ) Electron.s position : Position of electron s=r From equation of motion `r= V_(0)t+(1)/(2)(-(eE)/(M))t^(2)` `r= V_(0)t= (1)/(2)(eE)/(M) t^(2) hatj` Vector form `vecr= V_(0)t hati -(1)/(2) (eE)/(M) t^(2)hatj` |
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