A potential difference of 10, 000 V is applied across the capacitor and the charge
on the capacitor is 3.54 × 10−5 C. Compute the capacitance of the capacitor.
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12th
Physics
Electrostatic Potential and Capacitance
Effects of Dielectrics in Capacitors
A potential difference of 3...
PHYSICS
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Asked on December 30, 2019 by
Sreejala Vèrmâ
A potential difference of 300 V is applied between the plates of a plane capacitor spaced 1 cm apart. A plane parallel glass plate with a thickness of 0.5 cm and a plane parallel paraffin plate with a thickness of 0.5 cm are placed in the space between the capacitor plates find:
(i) intensity of electric field in each layer
(ii) the drop of potential in each layer
(iii) the surface charge density of the charge on the capacitor plates.
Given that: k
glass
=6, k
paraffin
=2
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ANSWER
CV
1
=3CV
2
....................(1)
V
1
+V
2
=300 ...................(2)
V
1
=75V
V
2
=225V
1. E
1
=
d
1
v
1
=
0.5
75×100
=1.5×10
4
V/m
E
2
=
d
2
v
2
=
0.5
225×100
=4.5×10
4
V/m
2. V
1
=75V
V
2
=225V
3. Q=
C
1
+C
2
C
1
C
2
V=
4
3
CV=
4
3
(
d
2ε
0
A
)×300
A
Q
=
4×0.5×10
−2
3×2×300×8.85×10
−12
=8×10
−7
C/m
2