A cloud layer of area 1.20 km² is 960 m above the Earth's surface. The Earth's surface and the cloud layer are considered as the plates of a parallel-plate capacitor. a) Calculate the capacitance of this capacitor. b) The air breaks down and conducts electricity if the electric field exceeds 3.00 x 106 V/m. Determine the maximum value of the charge the cloud can hold.

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Chapter8: Capacitance
Section: Chapter Questions
Problem 66AP: Suppose that the capacitance of a variable capacitor can be manually changed from 100 to 800 pF by...
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Can u solve number 3 part b plz
connected in parallel. Determine their capacitances.
2.
A parallel-plate capacitor consists of two identical
parallel conducting plates of sides 30.0 cm and 20.0 cm.
The two plates are separated by a distance d = 1.00 cm.
When this capacitor is charged under a voltage of 12.0 V, the
energy stored in the capacitor is 3.82 x 10-9 J.
a) Calculate the capacitance C of the capacitor.
b) Deduce the permittivity e of the dielectric material between
the plates.
and
a) D
b)
b= 30cm
Dielectric
a = 20cm
d = 1cm
3.
A cloud layer of area 1.20 km² is 960 m above the Earth's surface. The Earth's surface and the cloud
layer are considered as the plates of a parallel-plate capacitor.
a) Calculate the capacitance of this capacitor.
b) The air breaks down and conducts electricity if the electric field exceeds 3.00 × 10° V/m.
Determine the maximum value of the charge the cloud can hold.
4.
An air-filled parallel-plate capacitor is charged under a voltage of 74 V. When the capacitor becomes
fully charged, it is disconnected from the circuit. A dielectric of relative permittivity e, is then inserted
and completely fills the space between the plates. The voltage across the capacitor becomes 20 V.
Refer to table 1 (page 246) to identify the name of the inserted dielectric material.
5. /
An uncharged capacitor of capacitance C is connested in series
with a resistor of resistance R = 10 k2 across an idal battery of
emf E = 10 V, as shown in the adjacent figure. V.V and V3 are
three multimeters used as voltmeters, and A is a multimeter, used
Com
Com
V1
%3D
%3D
Transcribed Image Text:connected in parallel. Determine their capacitances. 2. A parallel-plate capacitor consists of two identical parallel conducting plates of sides 30.0 cm and 20.0 cm. The two plates are separated by a distance d = 1.00 cm. When this capacitor is charged under a voltage of 12.0 V, the energy stored in the capacitor is 3.82 x 10-9 J. a) Calculate the capacitance C of the capacitor. b) Deduce the permittivity e of the dielectric material between the plates. and a) D b) b= 30cm Dielectric a = 20cm d = 1cm 3. A cloud layer of area 1.20 km² is 960 m above the Earth's surface. The Earth's surface and the cloud layer are considered as the plates of a parallel-plate capacitor. a) Calculate the capacitance of this capacitor. b) The air breaks down and conducts electricity if the electric field exceeds 3.00 × 10° V/m. Determine the maximum value of the charge the cloud can hold. 4. An air-filled parallel-plate capacitor is charged under a voltage of 74 V. When the capacitor becomes fully charged, it is disconnected from the circuit. A dielectric of relative permittivity e, is then inserted and completely fills the space between the plates. The voltage across the capacitor becomes 20 V. Refer to table 1 (page 246) to identify the name of the inserted dielectric material. 5. / An uncharged capacitor of capacitance C is connested in series with a resistor of resistance R = 10 k2 across an idal battery of emf E = 10 V, as shown in the adjacent figure. V.V and V3 are three multimeters used as voltmeters, and A is a multimeter, used Com Com V1 %3D %3D
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