A uniform electric field exists between two parallel plates. The magnitude of th

Question

A uniform electric field exists between two parallel plates. The magnitude of the electric field if 50 N/C. an electron enters the field from the lower left corner, as shown. The initial velocity vector of the electron has a magnitude of 6 X 105 m/s and makes an angle =30 degrees with the x axis

let the x axis point to the right and the y axis point upward, from lower plate to the upper plate

a.) calculate the magnitude and direction of the electron's acceleration vector

b.) calculate the magnitude and direction of the electron's velocity vector at a time t=0.01s after it enters the electric field

Explanation / Answer

a)
Force on the electron, F = E*q = 50*1.6*10^-19 = 8*10^-18 N
so, acceleration, a = F/m = 8*10^-18/(9.1*10^-31) = 8.79*10^12 m/s2 <----magnitude
direction = towards the lower plate vertically downwards <-----direction

b)
we use the formula,
v = u + at
so, v = 6*10^5*(cos30(i) + sin30(j)) +(8.79*10^12(-j))*0.01 = 5.2*10^5(i) - 8.8*10^12(j)
so, magnitude = sqrt((5.2*10^5)^2 + (8.8*10^12)^2 ) = 8.8*10^12 m/s <----magnitude
direction = tan-inverse(8.8*10^12/5.2*10^5) = 90 degrees or towards the lower plate vertically downwards <---- direction

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