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1 year ago

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Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors a

re given equal but opposite uniform surface charge densities . Ignore edge effects and use Gauss's law to find for points far from the edges, the electric field between the plates. Find the field outside the plates on either side.

1 answer:

mote1985 [20]1 year ago

4 0

Answer:

E=0

Explanation:

Electric field due to each thin sheet of charge=\sigma/2\varepsilon

let us say the right plate has positive charge density \varepsilonand left sheet has a negative charge density -\varepsilon .

In the region between the plates,the electric field due to each plate is in same direction,

E=\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=\sigma/\varepsilon

in the region outside the plates, the field due to the plates is in opposite directions

E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=-\sigma/2\varepsilon+\sigma/2\varepsilon

E=0

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A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

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Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

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An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in th

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Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, $c=345J/^oC$

Initial temperature, $T_i=88^{\circ}C$

Final temperature, $T_f=45^{\circ}C$

We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :

$Q=mc\Delta T$

Let the mass of the object is 10 g or 0.01 kg

So,

$Q=0.01\times 345\times (88-45)$

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The weight of an object is the same on two different planets. The mass of planet A is only sixty percent that of planet B. Find

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Answer:

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Explanation:

The weight of an object on a planet is equal to the gravitational force exerted by the planet on the object:

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G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

$F_A=G\frac{M_Am}{R_A^2}$

For planet B,

$F_B=G\frac{M_Bm}{R_B^2}$

We also know that the weight of the object on the two planets is the same, so

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So we can write

$G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}$

We also know that the mass of planet A is only sixty percent that of planet B, so

$M_A = 0.60 M_B$

Substituting,

$G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}$

Now we can elimanate G, MB and m from the equation, and we get

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So the ratio between the radii of the two planets is

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