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Aliun [14]
3 years ago
10

Space vehicles traveling through Earth's radiation belts can intercept a significant number of electrons. The resulting charge b

uildup can damage electronic components and disrupt operations. Suppose a spherical metallic satellite 1.9 m in diameter accumulates 1.9 µC of charge in one orbital revolution. (a) Find the resulting surface charge density. (b) Calculate the magnitude of the electric field just outside the surface of the satellite, due to the surface charge.
Physics
1 answer:
Elena L [17]3 years ago
5 0

Answer:

a) 0.167 μC/m^2

b) 1.887 * 10^4 V/m

Explanation:

Hello!

First let's find the surface charge density:

a)

Since thesatellite is metallic, the accumalted charge will be uniformly distribuited on its surface. Therefore the charge density σ will be:

σ = Q/A

Where A is the area of the satellite, which is:

A=4πr^2 = πd^2 = π(1.9m)^2

Therefore:

σ = (1.9)/(π (1.9)^2) μC/m^2 = 0.167 μC/m^2

Now let's calculate the electric field

b)

Just outside the surface of the satellite the elctric field will be:

E = σ/ε0

Where      ε0=8.85×10^−12 C/Vm

Therefore:

E = (0.167*10^-6 C/m^2) / (8.85*10^-12  C/Vm) = 0.01887 *10^6 V/m

E = 1.887 * 10^4 V/m

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Answer:

The final velocity of the car is 36 m/s.

Explanation:

Given;

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the final velocity of the car is calculated as;

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v = 20 + (4 x 4)

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6 0
2 years ago
What is true of an object pulled inward in an electric field?
slava [35]

Answer:

option b

Explanation:

There is an object pulled inward in an electric field.

We have to find out of the four options given which is true.

a) The object has a neutral charge is false since when electric field pulls the object inward, there is a charge inside.

b) The object has a charge opposite that of the field, this option is correct since there will be an equal and opposite charge created by the object

c) The object has a negative charge will be correct only if the original charge was positive hence wrong

d) The object has a charge the same as that of the field is incorrect since this would be opposite the charge

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3 years ago
A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i
mars1129 [50]

Complete Question

The complete question is  shown on the first uploaded image  

 

Answer:

The electric field at that point is  E = 7500 \ N/C

Explanation:

From the question we are told that  

       The  radius of the inner circle is r_i  =  0.80  \ m

        The  radius of the outer circle is  r_o  =  1.20 \ m

       The  charge on the spherical shell q_n  =  -500nC  = -500*10^{-9} \ C

      The magnitude of the point charge at the center is  q_c =  + 300 nC  =  + 300 * 10^{-9} \ C

        The  position we are considering is  x =  0.60 m  from the center

Generally  the  electric field  at the distance x =  0.60 m  from the center  is mathematically represented as

                 E =  \frac{k *  q_c   }{x^2}

substituting values  

                  E =  \frac{k *  q_c   }{x^2}

where  k is  the coulomb constant with value k = 9*10^{9}  \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

     substituting values

                  E =  \frac{9*10^9  *  300 *10^{-9}}{0.6^2}

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3 years ago
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horsena [70]

Answer:

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Explanation:

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now we know that

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now we have

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P - 1500 = 6028 Pa

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Answer \|/

             

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