Answer :
Answer:
6.27μm
Explanation:
From the expression showing the relationship between the capacitance, Area and distance between the plates
C=EA/d..........1
Where E is the permittivity.
Also the equation showing the relationship between the voltage, charge and capacitance
C=q/v........... 2
If we equate..... 1 and..... 2 we have
EA/d=q/v
Making the distance 'd' subject of formula we arrive at
d=EAv/q
Also the charge is the product of the charge density and the Area I.e
q=σA
hence we have
d=εAv/σA
d=εv/σ
if we set ε=8.85*10⁻¹², v=170v and σ=24*10⁻⁵C/m
d=(8.85*10⁻¹²*170)/24*10⁻⁵
d=6.27*10⁻⁶m
d=6.27μm
The spacing between the two plates is 6.27μm
Answer:
0.0627 μcm or 6.27 x [tex]10^{-4}[/tex]cm
Explanation:
The charge (Q) on the plates can be expressed in terms of the surface charge density (σ) as follows;
Q = σA ---------------(i)
Where A is the overlap area between the plates
Also,
We know that capacitance (C) of a parallel plate capacitor is related to charge (Q) and potential difference (V) as follows;
Q = CV ------------(ii)
Combining the two equations,
CV = σA ----------------- (iii)
Also,
The capacitance (C) of a parallel plate capacitor is given by;
C = ε₀ x A / d --------------------(iv)
Where ε₀ = electric constant = 8.85 x [tex]10^{-14}[/tex]F/cm
d = distance (spacing) between the plates.
Substituting of C in equation(iv) into equation(iii) gives;
=> (ε₀ x A / d) x V = σ x A
A appears on both sides of the equation, so we can cancel it out.
=> ε₀ / d x V = σ
Making d the subject of the formula gives;
=> d = (ε₀ x V) / σ -------------------------(v)
Substitute the values of ε₀, V = 170V and σ = 24.0nC/[tex]cm^{2}[/tex] into the equation (v)
=> d = 8.85 x [tex]10^{-14}[/tex] x 170 / (24.0 x [tex]10^{-9}[/tex])
=> d = 0.0627 μcm or 6.27 x [tex]10^{-4}[/tex]cm
Therefore the spacing between the plates is d = 0.0627 μcm or 6.27 x [tex]10^{-4}[/tex]cm