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Bas_tet [7]
3 years ago
12

Ask Your Teacher In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C dir

ected downward. At 600 m above the ground, the electric field is 110 N/C downward. What is the average volume charge density in the layer of air between these two elevations?
Physics
1 answer:
strojnjashka [21]3 years ago
7 0

Answer:

1.475\times 10^{-13}\ C/m^3

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

A = Area

h = Altitude = 600 m

Electric flux through the top would be

-110A (negative as the electric field is going into the volume)

At the bottom

120A

Total flux through the volume

\phi=120-110\\\Rightarrow \phi=10A

Electric flux is given by

\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0

Charge per volume is given by

\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3

The volume charge density is 1.475\times 10^{-13}\ C/m^3

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Answer:

a.After 15 second Mr Comer's speed =  1.7 m/s

b.Distance travelled by Mr.Comer in 15 seconds  =   18.75\ m

Explanation:

a. Lets recall our first equation of motion V_{f} =V_{i} + at

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Plugging the values we have.

V_{f} =V_{i} + at

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b.

Lets find the distance and recall our third equation of motion.

V_{f} ^2-V{i}^2 = 2as

So s = distance covered.

Dividing both sides with 2a we have.

\frac{V_{f} ^2-V{i}^2}{2a} = 2as

Plugging the values.

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Hence, the time taken by the duck to cross the lake is, t= 4 s

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