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Please provide deep explaining of concepts, theories, and formulas that apply to each problem. I have a physics final coming up and I need to be prepared, these are the practice questions. thank you.

-12 points SerCP10 16.P025.WI Consider the Earth and a cloud layer 600 m above the planet to be the plates of a parallel-plate capacitor. (a) If the cloud layer has an area of 1.0 km-1000000 m, what is the capacitance? (b) If an electric field strength greater than 3.0 x 106 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold? (b) If an electric field strength greater than 3.0 × 106 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold? -/4 points SerCP10 16.P.033. Given a 1.50 F capacitor, a 3.75 F capacitor, and a 8.00 V battery, find the charge on each capacitor if you connect them in the following ways. (a) in series across the battery 1.50 F capacitor 3.75 F capacitor AC ALC (b) in parallel across the battery 1.50 F capacitor 3.75 F capacitor uC AAC

Explanation / Answer

Given that

Area(A) = 1.0*10^6 m^2

Distance(d) = 600 m

Maximum electric field strength Emax. = 3.0*10^6 N/C

The permitivity of the free space e0 = 8.85*10^-12 C^2/N.m^2

(a)
Capacitance (C) = e0*(A/d)

= (8.85*10^-12 C^2/N.m^2)*(1.0*10^6 m^2/600 m)

C = 1.47*10^-8 F

(b)
The maximum charge that the cloud can hold be obtained by applying

Q = C*(dealV)max.

Where V = E*d

Q = C*(Emax.*d)

Hence the maximum charge that the cloud can hold is

Q = C*(Emax.*d)

= (1.47*10^-8)*(3.0*10^6)*(600)

= 26.46 C

Given that
C1 = 1.50*10^-6 F
C2 = 3.75*10^-6 F
voltage V = 8.00 V

(a)
The equivalent capacitance is

1/C = 1/C1 + 1/C2

1/C = 1/( 1.50*10^-6 F) + 1/(3.75*10^-6 F)

C = 1.0714*10^-6 F

So the charge across each capacitance (1.50 µF and 3.75 µF) is

Q = C*V

= (1.0714*10^-6 F)*(8.00 V)

= 8.5712*10^-6 C

= 8.5712 C
(b)
The charge across capacitor 1.50 µF is

q1 =C1*V = (1.50*10^-6 F)*(8.00 V)
= 12*10^-6 C
= 24.75 C

The charge across capacitor 3.75 µF is
q2 = C2*V = (3.75*10^-6 F)*(8.00 V)
= 30*10^-6 C
= 30 C

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