Q.18 Two large conducting plate are placed parallel to each other with a separation of 2 cm between them. An electron starting from the rest near one of the plates reaches the other plate in 2 s. Find the Surface charge density of the plates
Dear Student ,
Mass of the electron: m = 9.1 x 10 -31 kg
Charge of the electron: e= -1.6 x 10 -19 C
Permittivity of vacuum: ε₀ =8.854 x 10 -12 F/m
d=0.02 m
t=2 x 10-6 s
We know that, F=ma
Again from electrostat we can write that , F=eE
Now ,
E = σ/ε₀
Therefore , F = eσ/ε₀
So , d=at2/2
or , a=2d/t2
Therefore, eσ/ε₀=2dm/t2
or , σ = 2dmε₀ /et2 [C/m2]
σ = (2 x 0.02 x9.1 x10-31 x 8.854 x 10-12) / 1.6 x 10-19 x (2 x 10-6)2
σ = 5.03 x 10-13 C/m2
.
Regards
Mass of the electron: m = 9.1 x 10 -31 kg
Charge of the electron: e= -1.6 x 10 -19 C
Permittivity of vacuum: ε₀ =8.854 x 10 -12 F/m
d=0.02 m
t=2 x 10-6 s
We know that, F=ma
Again from electrostat we can write that , F=eE
Now ,
E = σ/ε₀
Therefore , F = eσ/ε₀
So , d=at2/2
or , a=2d/t2
Therefore, eσ/ε₀=2dm/t2
or , σ = 2dmε₀ /et2 [C/m2]
σ = (2 x 0.02 x9.1 x10-31 x 8.854 x 10-12) / 1.6 x 10-19 x (2 x 10-6)2
σ = 5.03 x 10-13 C/m2
.
Regards