Answer :
Answer:
a. 11.5kv/m
b.102nC/m^2
c.3.363pF
d. 77.3pC
Explanation:
Data given
[tex]area=7.60cm^{2}\\ distance,d=2mm\\voltage,v=23v[/tex]
to calculate the electric field, we use the equation below
V=Ed
where v=voltage, d= distance and E=electric field.
Hence we have
[tex]E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m[/tex]
b.the expression for the charge density is expressed as
σ=ξE
where ξ is the permitivity of air with a value of 8.85*10^-12C^2/N.m^2
If we insert the values we have
[tex]8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2} \\102nC/m^{2}[/tex]
c.
from the expression for the capacitance
[tex]C=eA/d[/tex]
if we substitute values we arrive at
[tex]C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF[/tex]
d. To calculate the charge on each plate, we use the formula below
[tex]Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC[/tex]
For the air-filled capacitor:
(a) The electric field is 11.5 kV/m
(b) The surface charge density is 1.02×10⁻⁷C/m²
(c) The capacitance is 3.36pF
(d) Charge on the plate: 76pC
Capacitor:
Given that an air-filled capacitor consists of two parallel plates such that the:
Area of the plates, A = 7.6 cm² = 7.6×10⁻⁴ m²
distance between the plates, d = 2mm = 2×10⁻³m
voltage applied, V = 23V
(a) Electric field is given by:
E = V\d
E = 23/2×10⁻³
E = 11.5 kV/m
(b) The electric field for a parallel plate capacitor in terms of the surface charge density is given by:
E = σ/ε₀
where σ is the surface charge density:
σ = ε₀E
σ = 8.85×10⁻¹²×11.5×10³
σ = 1.02×10⁻⁷C/m²
(c) the capacitance is given by:
C = ε₀A/d
C = (8.85×10⁻¹²)×(7.6×10⁻⁴)/2×10⁻³
C = 3.36×10⁻¹²F
C = 3.36 pF
(d) The charge (Q) on the plate is given by:
Q = σA
Q = 1.02×10⁻⁷× 7.6×10⁻⁴
Q = 76 pC
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