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nadezda [96]
3 years ago
5

plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 85.0 nC . The plates are in vacuu

m. The electric field between the plates has a magnitude of 5.00×106 V/m . You may want to review (Pages 786 - 790) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Properties of a parallel-plate capacitor. Part A What is the potential difference between the plates?
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
3 0

Answer:

12500 V

Explanation:

The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:

\Delta V = E d

where

\Delta V is the potential difference between the plates

E is the electric field strength

d is the distance between the plates

For the capacitor in this problem, we have

E=5.00\cdot 10^6 V/m

d = 2.50 mm = 2.50\cdot 10^{-3} m

Substituting, we find

\Delta V = (5.00\cdot 10^6)(2.50\cdot 10^{-3})=12500 V

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Answer:

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4000 Joules (if you mean 2000 kilograms)

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An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s
gizmo_the_mogwai [7]

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

Q = ms\Delta T + mL

here we have

Q = 40(950)(680 - 32) + 40(450 \times 10^3)

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Q = 42624 kJ

Since this is the amount of aluminium per hour

so power required to melt is given by

P = \frac{Q}{t}

P = \frac{42624}{3600} kW

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Part B)

Total energy consumed by the furnace for 30 hours

Energy = power \times time

Energy = 13.93 kW\times 30 h

Energy = 417.9 kWh

now the total cost of energy consumption is given as

R = P \times 20 \frac{Cents}{kWh}

R = 417.9 kWh\times  20 \frac{cents}{kWh}

R = 8357.6 Cents

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