Question

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).

Answer 1

Answer:

Part a)

$Q = 3.26 \times 10^5 C$

Part b)

$E = 255.45 N/C$

Explanation:

Part a)

As we know that flux is defined as

$\phi = \frac{Q}{\epsilon_0}$

here we have

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

now we have

$3.69 \times 10^{16} = \frac{Q}{8.85 \times 10^{-12}}$

$Q = 3.26 \times 10^5 C$

Part b)

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

$R = 3.389\times 10^6 m$

now the electric field is given as

$E = \frac{kQ}{R^2}$

here we have

$E = \frac{(9\times 10^9)(3.26 \times 10^5)}{(3.389\times 10^6)^2}$

$E = 255.45 N/C$