Answer:
The diameters of the disk is 1.23 cm
Explanation:
Given information:
the space of two disks, d =0.5 mm = 0.0005 m = 5 x [tex]10^{-4}[/tex] m
the transferred electron, n = 1.70 x [tex]10^{9}[/tex]
electric field strength, E = 2.6 x [tex]10^{5}[/tex] N/C
to find the diameter of the disk we can use the following equation
A = πD²/4 ...........................................(1)
where
D = the distance of the disk
A = the area of the disk
first, we have to find the are of the disk using the capacitance equation
C = ε₀A/d...........................................(2)
A = Cd/ε₀ where C = Q/V (Q is total charge and V is potential difference)
thus
A = Qd/Vε₀.........................................(3)
now substitute V = Ed and Q = ne, so
A = (ned)/(Edε₀)
= ne/Eε₀..........................................(4)
e = 1.6 x [tex]10^{-19}[/tex] C
now can substitute equation (4) to the first equation
A = πD²/4
D² = 4A/π
D = √4A/π
= √(4ne)/(πEε₀) , ε₀ = 9.85 x [tex]10^{-12}[/tex] C²/Nm²
= √4(1.70 x [tex]10^{9}[/tex])(1.6 x [tex]10^{-19}[/tex])/π(2.6 x [tex]10^{5}[/tex] )(8.85 x [tex]10^{-12}[/tex] )
= 0.0123 m
= 1.23 cm
