The ink drops have a mass = 1.00×10-11 each and leave the nozzle and travel hori

Question

The ink drops have a mass = 1.00×10-11 each and leave the nozzle and travel horizontally toward the paper at velocity = 25.0 . The drops pass through a charging unit that gives each drop a positive charge by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length = 1.95 , where there is a uniform vertical electric field with magnitude = 8.50×104 .
If a drop is to be deflected a distance = 0.340 by the time it reaches the end of the deflection plate, what magnitude of charge must be given to the drop? Assume that the density of the ink drop is 1000 , and ignore the effects of gravity.

Explanation / Answer

when an electron enters the electric field it has velocity v in x- direction. the electron pointing vertically upward imparts a uniform acceleration to the electron                  a = F / m                        = - E q / m where q = charge on ink drop              m = mass of ink drop = 1*10^ -11 kg              E  = electric field between between the plates = 8.05*10^ 4 N /C                    The electron vertical position                y = ( 1 /2 ) a t^2                    = ( 1 /2 ) (- Eq / m ) t^2   ........(1)               sin ce horizontal component of acceleration is constant .the horizonatl position is                x = v   t                t = x / v   v= velocity of drop = 25 m/s substitute t value in equation                  y =  [ - Eq / 2m (v)^2 ] x^2        where y = deflected distance by drop=0.34 mm               x = length of the plates= 1.95 cm             0.34*10^ -3 m = [(8.05*10^ 4 N /C) q / ( 1*10^ -11 )(25 ^2 ) ] (0.0195) ^2                       q = (0.894)( 1*10^ -11 )(25 ^2 ) / (8.05*10^ 4 N /C)                         = 6.94*10^ -14 C         when an electron enters the electric field it has velocity v in x- direction. the electron pointing vertically upward imparts a uniform acceleration to the electron                  a = F / m                        = - E q / m where q = charge on ink drop              m = mass of ink drop = 1*10^ -11 kg              E  = electric field between between the plates = 8.05*10^ 4 N /C                    The electron vertical position                y = ( 1 /2 ) a t^2                    = ( 1 /2 ) (- Eq / m ) t^2   ........(1)               sin ce horizontal component of acceleration is constant .the horizonatl position is                x = v   t                t = x / v   v= velocity of drop = 25 m/s substitute t value in equation                  y =  [ - Eq / 2m (v)^2 ] x^2        where y = deflected distance by drop=0.34 mm               x = length of the plates= 1.95 cm             0.34*10^ -3 m = [(8.05*10^ 4 N /C) q / ( 1*10^ -11 )(25 ^2 ) ] (0.0195) ^2                       q = (0.894)( 1*10^ -11 )(25 ^2 ) / (8.05*10^ 4 N /C)                         = 6.94*10^ -14 C        

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