Question

An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between two parallel plates, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 10^2 N/C and separation between the charged plates is 2.0 cm. a.) Determine the horizontal distance traveled by the electron when it hits the plate. b.)Determine the velocity of the electron as it strikes the plate.

Answer 1

Answer:

Explanation:

Given that  

speed u=4×10⁶ m/s

electric field E=4×10² N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates

acceleration a=qE/m=1.6×10⁻¹⁹×4×10²/9.1×10⁻³¹=0.7×10¹⁴=7×10¹³ m/s

now we find the horizantal distance travelled by electrons hit the plates

horizantal distance X=u[2y/a]^1/2

=4×10⁶[2×2×10⁻²/7×10¹³]^1/2

=9.5cm

now we find the velocity f the electron strike the plate

v²-(4×10⁶)²=2×7×10¹³×2×10⁻²

v²=16×10¹²+28×10¹¹

v²=1.88×10¹³m/s

speed after hits =>V=4.34×10⁶ m/s