Question

xt-4.00 cm

An electron with a kinetic energy of 2.00 keV (1 eV = 1.602 × 10^-19 J) (See the figure above.) is fired horizontally across a horizontally oriented charged conducting plate with a surface charge density of σ = +4.00×10^-6 C/m². Taking the positive direction to be upward (away from the plate), what is the vertical deflection of the electron after it has traveled a horizontal distance of 4.00 cm?

Show all work. Include your explanation and show your calculation in detail.

Answer 1

The charged plate will give force on the electron due to the electric field it produces. Electric field, E = 6/2 elementof_0 [Using Gauss's Law] = 4 times 10^-6/2 times 8.85 times 10^-12 v/ m implies E = 0.225 times 10^6 V/m Electric field is along positive Y - axis Therefore Force on the moving electron F = -eE (F is opposite to E) = D F = -1.6 times 10^-19 times 0.225 times 10^6 is Newton = D F = -0.36 times 10^-13 is Newton So, the force is in the negative Y axis direction If we consider that electric fields lines are parallel and it gives a constant force in the downward direction then we can consider the motion of electron as a protective motion \r\n Now, u middot E = E = 2 Kev = D E = 2 times 10^3 times 1.6 times 10^-19 J = 3.2 times 10^-16 J Therefore 1/2 mv^2_0 = 3.2 times 10^-16