Recent content by nks27

sorry for late post. substituting v_x = vcos60 and v_y = vsin60 q= 1.6 X 10^-19 and V= 100 V i get v^2 to be 1.41 X 10^14 using it to calculate the k.E doesn't give the right answer :( the " q/(4πεₒr) " was for electric potential .

i the use this formula work done against electric = K.E + P.E field K.E = work done - P.E = qV - q/(4πεₒr) but i don't get the answer. Cud u please correct my careless mistake if there's any?

Homework Statement Two metal plates, PQ AND RS are separated by a distance of 15 mm. PQ maintained at potential of +100V relative to RS. Beam of electrons of different kinetic energies directed a slit ,on plate PQ, at angle of 60 degrees to plate. To find the K.E of electrons that 'just'...