a parallel plate capacitor as an area of 4cm^2& plate sepretion of 2mm calculate its capacitance
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1
Given,
d=2mm
v=400v
t=10
−6
A=60cm
2
at t=0 no charge on the capacitor, so potential drop across the capacitor =0
at t=10
−6
sec capacitor get fully charge to potential drop across capacitor=400v
I
D
=ε
o
dt
dϕ
=ε
o
dt
d
(EA)=ε
o
A
dt
dE
=
dt
ε
o
A
(
dt
dv
)
=
d
ε
o
A
Δt
Δv
=
d
ε
o
A
(
f
f
−t
i
v
f
−v
i
)
=
2×10
−3
8.85×10
−12
×60×10
−4
(
10
−6
−0
400−0
)
=
2×10
−3
8.85×10
−12+6
×60×10
−4
×400
=3×8.85×10
I
D
=12×8.85×10
−4
I
D
=1.06×10
2
A.
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