Question

(3) An electron is projected with horizontal velocity v.-5x10s m/s into a uniform electric field of magnitude 4000 N/C pointed upwards that is presented over a horizontal distance of 2 cm (a) Find the magnitude and direction of the force due to the electric field. Is it justifiable to ignore the gravitational force?

Answer 1

As we know that F = qE = -1.6 times 10^-19 times 4000 = -6.4 times 10^-16 N Magnitude = Vertical-bar F Vertical-bar = 6.4 times 10^-16 N rightarrow In downward direction As we know F_g = mg = 9.1 times 10^-31 times 9.8 = 89.18 times 10^-31 N since gravitational force is very small in comparison to electrostatic force hence we can neglect gravitational force time spent in electric field t = d/v = .02/5 times 10^6 = 4 times 10^-9 s acceleration of electron a = F/m = 6.4 times 10^-16/9.1 times 10^-31 = 7.03 times 10^14 m/s^2 therefore vertical distance travelled y = 1/2 at^2 = 1/2 times 7.03 times 10^14 times (4 times 10^-9)^2 y = 5.624 times 10^-3 m tan theta = y/x = 5.625 times 10^-3/.02 = > theta = 15.7 degree clockwise from positive axis