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50 μC of negative charge is placed on an insulating pith ball and lowered into a insulating plastic container, suspended from an insulating thread attached to the lid of the box. After the box is entirely sealed, the electric flux through the sides of the box is:_______

a. 5. 65 Times 10^6 N m^2/C.
b. 5. 65 Times 10^5 N m^5/C.
c. -5. 65 Times 10^6 N m^2/C.
d. 50 x 10^-6 N m^2/C.
e. -5.65 Times 10^5 N m^2/C.
f. Can't tell unless the dimensions of the box are given.

Respuesta :

Answer:

c. [tex]-5. 65 \times 10^6 N m^2/C.[/tex]

Explanation:

The calculation of the electric flux through the sides of the box is shown below:-

Negative charge in insulating pitch ball, [tex]q = 50\times 10^{-6}[/tex]

[tex]Permittivity = 8.854 \times 10^{-12} F/m[/tex]

Now, we are placing the values into the formula which is here below:-

[tex]Flux = \frac{Negative\ charge}{Permittivity}[/tex]

[tex]= \frac{50\times 10^{-6}}{8.854 \times 10^{-12}}[/tex]

= [tex]-5. 65 \times 10^6 N m^2/C.[/tex]

Therefore we divided the negative charge by permittivity to reach out the electric flux through the sides of the box.

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