Answer:
voltage between the plates is 4.952 × [tex]10^{-26}[/tex] V
Explanation:
given data
plate separated distance = 2.67 cm
electron speed = 1.32 × [tex]10^{7}[/tex] m/s
solution
we will get here first force that is express as
force in parallel plate F = [tex]\frac{eV}{d}[/tex] ..............1
and force by Newton second law F = ma .............2
equate equation 1 and 2
ma = [tex]\frac{eV}{d}[/tex] .................3
and here we know as kinematic equation
v²- u² = 2 × a × s ...........4
so for initial speed acceleration will be
a = [tex]\frac{v^2-u^2}{2\times s}[/tex]
a = [tex]\frac{(1.32 \times 10^7)^2}{2\times 2.67 \times 10^{-2}}[/tex]
a = 3.262 × [tex]10^{-13}[/tex] m/s²
now we put a in equation 3 and we get v
ma = [tex]\frac{eV}{d}[/tex]
9.1093 × [tex]10^{-31}[/tex] × 3.262 × [tex]10^{-13}[/tex] = [tex]\frac{1.602 \times 10^{-19} V}{2.67 \times 10^{-2}}[/tex]
solve it we get
v = 4.952 × [tex]10^{-26}[/tex] V
