Answer:
The charge inside the cube is null.
Explanation:
If we apply the gauss theorem with a cubical gaussian surface of the size of the cube:
![\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\frac{q_{in}}{\varepsilon_{0}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Coint_%7BS%7D%20%5Cvec%7BE%7D%5C%2C%5Cvec%7Bds%7D%3D%5Cfrac%7Bq_%7Bin%7D%7D%7B%5Cvarepsilon_%7B0%7D%7D)
If we consider than the direction of the electric field is
, we can solve the problem differentiating the integral for each face of the cube:
![\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} \vec{E}\,\vec{ds_1}+\displaystyle\int_{S_2} \vec{E}\,\vec{ds_2}+\displaystyle\int_{S_3} \vec{E}\,\vec{ds_3}+\displaystyle\int_{S_4} \vec{E}\,\vec{ds_4}+\displaystyle\int_{S_5} \vec{E}\,\vec{ds_5}+\displaystyle\int_{S_6} \vec{E}\,\vec{ds_6}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Coint_%7BS%7D%20%5Cvec%7BE%7D%5C%2C%5Cvec%7Bds%7D%3D%5Cdisplaystyle%5Cint_%7BS_1%7D%20%5Cvec%7BE%7D%5C%2C%5Cvec%7Bds_1%7D%2B%5Cdisplaystyle%5Cint_%7BS_2%7D%20%5Cvec%7BE%7D%5C%2C%5Cvec%7Bds_2%7D%2B%5Cdisplaystyle%5Cint_%7BS_3%7D%20%5Cvec%7BE%7D%5C%2C%5Cvec%7Bds_3%7D%2B%5Cdisplaystyle%5Cint_%7BS_4%7D%20%5Cvec%7BE%7D%5C%2C%5Cvec%7Bds_4%7D%2B%5Cdisplaystyle%5Cint_%7BS_5%7D%20%5Cvec%7BE%7D%5C%2C%5Cvec%7Bds_5%7D%2B%5Cdisplaystyle%5Cint_%7BS_6%7D%20%5Cvec%7BE%7D%5C%2C%5Cvec%7Bds_6%7D)
![\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} E_0\hat{x}\,\hat{x}ds_1+\displaystyle\int_{S_2} E_0\hat{x}\,\hat{-x}ds_2+\displaystyle\int_{S_3} E_0\hat{x}\,\hat{y}ds_3+\displaystyle\int_{S_4} E_0\hat{x}\,\hat{-y}ds_4+\displaystyle\int_{S_5} E_0\hat{x}\,\hat{z}ds_5+\displaystyle\int_{S_6} E_0\hat{x}\,\hat{-z}ds_6](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Coint_%7BS%7D%20%5Cvec%7BE%7D%5C%2C%5Cvec%7Bds%7D%3D%5Cdisplaystyle%5Cint_%7BS_1%7D%20E_0%5Chat%7Bx%7D%5C%2C%5Chat%7Bx%7Dds_1%2B%5Cdisplaystyle%5Cint_%7BS_2%7D%20E_0%5Chat%7Bx%7D%5C%2C%5Chat%7B-x%7Dds_2%2B%5Cdisplaystyle%5Cint_%7BS_3%7D%20E_0%5Chat%7Bx%7D%5C%2C%5Chat%7By%7Dds_3%2B%5Cdisplaystyle%5Cint_%7BS_4%7D%20E_0%5Chat%7Bx%7D%5C%2C%5Chat%7B-y%7Dds_4%2B%5Cdisplaystyle%5Cint_%7BS_5%7D%20E_0%5Chat%7Bx%7D%5C%2C%5Chat%7Bz%7Dds_5%2B%5Cdisplaystyle%5Cint_%7BS_6%7D%20E_0%5Chat%7Bx%7D%5C%2C%5Chat%7B-z%7Dds_6)
E₀ is a constant and each surface is equal to each other, so: ![S_1=S_2=S_i=S](https://tex.z-dn.net/?f=S_1%3DS_2%3DS_i%3DS)
Therefore:
![\displaystyle\oint_{S} \vec{E}\,\vec{ds}=E_0\displaystyle\int_{S_1} \,ds_1+E_0\displaystyle\int_{S_2} -1\,ds_2+0+0+0+0=E_0S-E_0S=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Coint_%7BS%7D%20%5Cvec%7BE%7D%5C%2C%5Cvec%7Bds%7D%3DE_0%5Cdisplaystyle%5Cint_%7BS_1%7D%20%5C%2Cds_1%2BE_0%5Cdisplaystyle%5Cint_%7BS_2%7D%20-1%5C%2Cds_2%2B0%2B0%2B0%2B0%3DE_0S-E_0S%3D0)
![\displaystyle\oint_{S} \vec{E}\,\vec{ds}=0=\frac{q_0}{\varepsilon_0} \longleftrightarrow q_0=0c](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Coint_%7BS%7D%20%5Cvec%7BE%7D%5C%2C%5Cvec%7Bds%7D%3D0%3D%5Cfrac%7Bq_0%7D%7B%5Cvarepsilon_0%7D%20%5Clongleftrightarrow%20q_0%3D0c)
Answer: Distance
I’m not sure if it is correct, but I am pretty sure it is......
I hope this helps :)
<span> Balanced. If not so, then the car cannot move with uniform speed. The frictional force is balanced by the force exerted by the engine. </span>
The speed of sound is influenced by several factors, including medium, density and temperature. The rate at which sound waves moves varies widely from one situation to the next and can change dramatically in a short period of time.