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# A ball of mass 2kg, charge 1×10-6 C is dropped from top of a high tower . In space electric field exists in horizontal direction away from the tower whoch vareis as E=(5-2x)×106 V/m. Find the maximum horizontal distance ball can go from the tower and displacement of ball at this instant from the point of projection?

12 years ago

hi ,

look the figure ,below . it shows only how the horizonta motion occurs

dv/dt = (5-2x)×106   ( 1×10-6   )

=(5-2x)

dx/dt = v

dv/dx = (5-2x)/v

v dv = (5-2x) dx

v^2/ 2]  from 0 to v =  5x - x^2 ]  from o to x

v^2/ 2  =  5x - x^2

for v =0 ,

5x - x^2 = 0

x= 0,  5

so ball will move back at x= 5 , this will be max. horizontal distance

displacement of ball at this instant from the point of projection ,

dx/ dt = v = √ 10x -  2 x^2

from here we can get time when it reaches  x= 5 ,

in this duration vertical displacement =  1/2 g t ^2

total  displacement  =  √ ( horiz. displacemen ) ^2 + ( vertical displacement  ) ^2 