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Monica [59]
3 years ago
13

A parallel plate capacitor is created by placing two large square conducting plates of length and width 0.1m facing each other,

separated by a 1 cm gap with vaccum between the plates. We connect the capacitor to power supply, charge it to a potential difference VO 5 kV, and disconnect the power supply. We then insert a sheet of insulating plastic material between the plates, completely filling the space between them. We find that the potential difference decreases to 1 kV while the charge on each capacitor plate remains constant. Find the original capacitance CO. 885pf 17.7pf 8.85pf 1.77pf
Physics
1 answer:
borishaifa [10]3 years ago
5 0

Answer:

8.854 pF

Explanation:

side of plate = 0.1 m ,

d = 1 cm = 0.01 m,

V = 5 kV = 5000 V

V' = 1 kV = 1000 V

Let K be the dielectric constant.

So, V' = V / K

K = V / V' = 5000 / 1000 = 5

C = ε0 A / d = 8.854 x 10^-12 x 0.1 x 0.1 / 0.01 = 8.854 x 10^-12 F

C = 8.854 pF

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Answer:

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b) 3.466 × 10¹¹ N/C

Explanation:

a)

p(r) = -A exp ( - 2r/a₀)

Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV  =  -A  ₀∫^∞ ₀∫^π ₀∫^2π   exp ( - 2r/a₀)r² sinθdrdθd∅

Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e

now using integration by parts;

A = e / πa₀³

p(r) =  - (e / πa₀³) exp (-2r/a₀)

Now Net charge inside a sphere of radius a₀ i.e Qnet is;

= e - (e / πa₀³)  ₀∫^a₀ ₀∫^π ₀∫^2π  r² exp (-2r/a₀)dr

= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b)

Using Gauss's law,

E × 4πa₀ ² = Qnet / ∈₀

E = 4πa₀ ² × Qnet × 1/a₀²

E = 3.466 × 10¹¹ N/C

4 0
3 years ago
a car goes from a velocity zero to a velocity of 15 meters per second East in 2.1 seconds. What is the car's acceleration?
leva [86]
31.5 would be the acceleration.
3 0
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Answer:

Energy expenditure in K cals/min = 10 K cals /min (approximately)

Explanation:

As we know

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Weight of Jazz= 172lb=78.02kg

putting the values in formula,

Energy expenditure in K cals/min=  7.6 x 3.5 x 78.02 / 200

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Therefore, Energy expenditure in K cals/min by Jazz will be approximately 10 K cals /min

8 0
3 years ago
A certain ideal gas has molar heat capacity at constant volume CV. A sample of this gas initially occupies a volume V0 at pressu
ANTONII [103]

Answer:

Explanation:

The processes are described on the image attached below. The isobaric process consists of an horizontal line, the adiabatic expansion is described by a polytropic curve:

P_{2} \cdot V_{2}^{\gamma} = P_{3} \cdot V_{3}^{\gamma}

Where:

\gamma = \frac{c_{p}}{c_{v}}

\gamma = 1 + \frac{R}{c_{v}}

Final pressure is:

P_{3} = P_{2}\cdot \left(\frac{V_{2}}{V_{3}}  \right)^{\gamma}

P_{3} = P_{o}\cdot \left(\frac{1}{2}\right)^{\gamma}

P_{3} = \frac{P_{o}}{2^{\gamma}}

8 0
3 years ago
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