1.

A charged particle of charge `2 muC` and mass `10` milligram, moving with a velocity of `1000 m//s` enters a uniform electric field of strength `10^(3) N//C` directed perpendicular to its direction of motion. Find the velocity and displacement of the particle after `10s`.

Answer» Here, `q = 2 muC = 2xx10^(-6) C`,
`m = 10 mg = 10^(-5) kg`
`v_(x) = 1000 m//s E_(y) = 10^(3) N//C `
`v_(y) = (u_(y) + a_(y)t = 0 + (qE_(y) t)/(m))`
`(2xx10^(-6)xx10^(-3)xx10)/(10^(-5)) = 2000 m//s`
As ` v_(x)` remains constant, therefore net velocity after `10s`
`= sqrt(V_(x)^(2) + v_(y)^(2)) = sqrt((1000)^(2) + (2000)^(2))`
`= 1000 sqrt(5) m//s`
Displacement along x-axis after `10s`
`x = v_(x) t = 1000xx10 = 10000 m`
Displacement along y-axis after `10s`
`y = u_(y) xx t+ (1)/(2) (q E_(y))/(m) t^(2)`
`= 0+ (1)/(2) xx (2xx10^(-6)xx10^(3))/(10xx10^(-6)) (10^(2)) = 10000m`
Net displacement `= sqrt(x^(2) + y^(2))`
`= sqrt((10000)^(2) + (10000)^(2)) = 10000 sqrt(2) m`


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