Question

Measurements indicate that there is an electric field surrounding the Earth. Its magnitude is about 150 N/C at the Earth's surface and points inward toward the Earth's center. What is the magnitude of the electric charge on the Earth? Express your answer using two significant figures.

Answer 1

The charge of earth is
`-6.8* 10^(5)Columbs`

Step-by-step explanation:

Assuming earth as a spherical body we have

For a sphere of radius 'r' and charge 'q' the electric field generated at a distance 'r' form the center of sphere is given by the equation

`E=(1)/(4\pi \epsilon _o )\cdot (Q)/(r^(2))`

where

'Q' is the total charge on sphere

Now at a distance 'r' equal to radius of earth(6371 km) we have the electric field strength is 150N/C

Using the given values we obtain

`150=(1)/(4\pi \epsilon _o)(Q)/((6371* 10^(3))^2)\\\\\therefore Q=150* (6371* 10^(3))^(2)* 4\pi \epsilon _o\\\\\therefore Q=6.8* 10^(5)Columbs`

Now since the electric field is inwards thus we conclude that this charge is negative in magnitude.

Answer 2

Magnitude of electric field on Earth = Q = 6.8 × 10⁵ C

Step-by-step explanation:

Electric field = E = 150 N/C

Distance from the center of the earth to the surface = Radius of the earth

Radius of the earth = R = 6.38× 10⁶ m

E = k Q / R² is the basic formula for the electric field. k = 9 × 10⁹ N m²/C²

150 = (9 × 10⁹)(Q) / (6.38× 10⁶ )²

⇒ Charge = Q = (150)(6.38× 10⁶ )²/(9 × 10⁹)

= 6.8 × 10⁵ C(2 significant figures).