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Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of 3.69×1016 N⋅m2/C into the planet's surface.Part A: Calculate the total electric charge on the planet. Part B: Calculate the magnitude of the electric field at the planet's surface (refer to the astronomical data inside the back cover).

Respuesta :

Answer:

Part a)

[tex]Q = 3.26 \times 10^5 C[/tex]

Part b)

[tex]E = 255.45 N/C[/tex]

Explanation:

Part a)

As we know that flux is defined as

[tex]\phi = \frac{Q}{\epsilon_0}[/tex]

here we have

[tex]\phi = 3.69 \times 10^{16} N.m^2/C[/tex]

now we have

[tex]3.69 \times 10^{16} = \frac{Q}{8.85 \times 10^{-12}}[/tex]

[tex]Q = 3.26 \times 10^5 C[/tex]

Part b)

Also we know that the radius of the planet is given as

[tex]R = 3.389\times 10^6 m[/tex]

now the electric field is given as

[tex]E = \frac{kQ}{R^2}[/tex]

here we have

[tex]E = \frac{(9\times 10^9)(3.26 \times 10^5)}{(3.389\times 10^6)^2}[/tex]

[tex]E = 255.45 N/C[/tex]

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