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posledela
3 years ago
8

A negative charge -Q is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a c

onducting wire between it and the earth. Is any excess charge induced on the inner surface of the metal? Is there any excess charge on the outside surface of the metal? Why or why not? Would someone outside the solid measure an electric field due to the charge -Q? Is it reasonable to say that the grounded conductor has shielded the region outside the conductor from the effects of the charge -Q? In principle, could the same thing be done for gravity? Why or why not?
Physics
1 answer:
tester [92]3 years ago
3 0

Answer:

a)  + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) E=0

Explanation:

Let's analyze the situation when a negative charge is placed inside the cavity, it repels the other negative charges, leaving the necessary positive charges to compensate for the -Q charge. The electrons that migrated to the outer part of the sphere, as it is connected to the ground, can pass to the earth and remain on the planet; therefore on the outside of the sphere the net charge remains zero.

With this analysis we can answer the specific questions

a)  + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) If we create a Gaussian surface on the outside of the sphere the net charge on the inside of this sphere is zero, therefore there is no electric field, on the outside

d) If it is very reasonable and this system configuration is called a Faraday Cage

e) We cannot apply this principle to gravity since there are no particles that repel, in all cases the attractive forces.

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A spherical shell has an outside radius of 2.60 cm and an inside radius of a. The shell wall has uniform thickness and is made o
lbvjy [14]

Answer:

The answer is below

Explanation:

a) The volume of a sphere is:

Volume = (4/3)πr³; where r is the radius of the shell.

Given the outside radius of 2.60 cm and inner radius of a cm, the volume of the spherical shell is:

Volume of spherical shell = \frac{4}{3} \pi (2.6^3-a^3) cm³

But Density = mass / volume;   Mass = density * volume.

Therefore, mass of spherical shell = density * volume

mass of spherical shell = 4.70\ g/cm^3 * \frac{4}{3} \pi (2.6^3-a^3) cm³

Mass of liquid = volume of inner shell * density of liquid

Mass of liquid = \frac{4}{3} \pi a^3\ cm^3*1.23\ g/cm^3

Total mass of sphere including contents = mass of spherical shell + mass of liquid

Total mass of sphere including contents (M) = 4.70\ g/cm^3 * \frac{4}{3} \pi (2.6^3-a^3)\ cm^3  +  \frac{4}{3} \pi a^3\ cm^3*1.23\ g/cm^3 =

Total mass of sphere including contents (M) = (346 - 14.5a³) grams

b) The mass is maximum when the value of a = 0

M = 346 - 14.5a³

M' = 43.5a² = 0

43.5a² = 0

a = 0

4 0
3 years ago
Can someone check my answers pls
o-na [289]
Hmmm... I’d say that’s pretty accurate. Good job! Well wishes on whatever it is you’re doing.
7 0
4 years ago
3. Calculate the wavelength of wave that has a frequency of 4.75 x 1012Hz.
Klio2033 [76]
  • Frequency=v=4.75×10^12Hz
  • Wavelength=?

We know

\boxed{\sf \lambda=\dfrac{C}{V}}

\\ \sf\longmapsto \lambda=\dfrac{3\times 10^8ms^{-1}}{4.75\times 10^{12}s^{-1}}

\\ \sf\longmapsto \lambda=0.631\times 10^{-4}m

\\ \sf\longmapsto \lambda=6.31\times 10^{-5}m

4 0
3 years ago
Which of the following would have an effect on an object's mass?
ICE Princess25 [194]
None of the above

mass is measurement of how much stuff in inside something.
if you freeze or heat an object, you merely change the state of the object. the mass is conserved.
if you change the elevation, nothing happens to the mass. the stuff will not leave the object.
so it is None of the above


8 0
3 years ago
Find the mass of the car.
SpyIntel [72]

\huge\boxed{770.8\overline{3}\ \text{kg}}

We know that force equals mass times acceleration, so substitute in the known values and solve.

\begin{aligned}F&=ma\\1850&=m*2.4\\1850\div2.4&=m*2.4\div2.4\\\boxed{770.8\overline{3}}&=m\end{aligned}

4 0
3 years ago
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