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26 March, 11:45

The Earth's surface, on average, carries a net negative charge (while the clouds and lower atmosphere carry a net positive charge of nearly equal magnitude), giving rise to an electric field at its surface which is about 150 N/C in size. 1 Assuming a spherical Earth, determine (a) the total charge on Earth's surface responsible for this field and the corresponding surface charge density. (b) Given the 150 N/C field at Earth's surface, how much lower is the field strength right below a cloud at a height of 5 km? [Careful for rounding errors here.]

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  1. 26 March, 11:50
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    (a) Q = - 6.765 * 10⁵ C; σ = 1.33 * 10⁻⁹ C/m²

    (b) 0.23 N/C

    Explanation:

    (a) Electric field at the surface of a sphere is given as:

    E = kQ/r²

    Where

    k = Coulombs constsnt

    Q = charge

    r = radius of sphere

    To find charge Q, we make Q subject of the formula:

    Q = (E * r²) / k

    Hence, charge, Q, at the surface of the earth, having radius, r = 6.371 * 10⁵ and electric field, E = 150 N/C is:

    Q = [150 * (6.371 * 10⁶) ²] / (9 * 10⁹)

    Q = 6.765 * 10⁵ C

    Since we're told that the charge at the earth's surface is negative,

    Q = - 6.765 * 10⁵ C

    Surface charge density, σ, given as:

    σ = |Q|/A

    Where

    |Q| = magnitude of charge

    A = surface area.

    Surface area, A, of the earth is given as:

    A = 4πr²

    A = 4π * (6.371 * 10⁶) ²

    A = 510064471909788 m²

    σ = 6.765 * 10⁵/510064471909788

    σ = 1.33 * 10⁻⁹ C/m²

    (b) At a height 5km from the earth's surface, the electric field will be:

    E = kQ / (r + 5km) ²

    r + 5km = 6376km = 6.376 * 10⁶m

    => E = (9 * 10⁹ * 6.765 * 10⁵) / (6.376 * 10⁶) ²

    E = 149.77 N/C

    The difference between the electric field at the surface of the earth and at a height of 5km is:

    159 - 149.77 N/C = 0.23 N/C
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