A parallel plate capacitor has plate area of 25cm² and a seperation of 2.0mm between it's plate. The capacitor is connected to 12volt battery. find the charge on the capaciter.
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Plate area A=25cm
2
=2.5×10
−10
J
Separation d=2mm=2×10
−3
m
Potential v=12v
We know
C=
d
ε
0
A
=
2×10
−3
8.85×10
−12
×2.5×10
−3
=11.06×10
−12
F
Capacitance is also written as:
C=
V
q
Equating the two values:
⇒11.06×10
−12
=
12
q
1
⇒q
1
=1.32×10
−10
C
Then d= decreased to 1mm
d=1mm=1×10
−3
m
C=
d
ε
0
A
=
v
q
=
1×10
−3
8.85×10
−12
×2.5×10
−3
=
12
2
⇒q
2
=8.85×2.5×12×10
−12
=2.65×10
−10
C.
∴ The extra charge given to plate =(2.65−1.32)×10
−10
=1.33×10
−10
C
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