Question

On a clear day the electric field in the atmosphere near the earth's surface is 100 N/C, directed vertically downward. (a) If we adopt the convention that the potential at the earth's surface is 0, what is the potential 492 m above the surface? (b) What is the potential at the top of Pike's Peak, 14,110 ft above sea level? (Hint: Do not assume that the surface in part (a) is at sea level.) Explain.

Answer 1

a) Given E = 100 N/C = 100 V/m downward.;
Vearths surface = 0; Δy = 492 m.

Direction of the E at any point is the direction in which a free positive charge tends to move, this also down the potential gradient. So, E is downward means, a lower point is at a lower pot. and a higher point is at a higher potential.

The relation between E and change in pot, ΔV is given by the eqn,
E = - ΔV/Δy in one dimension.
Therefore, ΔV = -E.Δy = - -100x492 = +49200 V = +49.2 kV.
(E is -ve as we up; y is +ve).
Or as explained earlier, pot is higher at higher points.
Therefore, the pot at the given point P, 492 m above earth's surface,
VP = 0 + 49.2 kV = 49.2 kV.

b) Given E = 100 N/C = 100 V/m downward.;
Vearths surface = 0; Δy = 14,110 ft = 4300.728 m

Direction of the E at any point is the direction in which a free positive charge tends to move, this also down the potential gradient. So, E is downward means, a lower point is at a lower pot. and a higher point is at a higher potential.

The relation between E and change in pot, ΔV is given by the eqn,
E = - ΔV/Δy in one dimension.
Therefore, ΔV = -E.Δy = - -100x4300.728 = +430072.8 V = +430.0728 kV.
(E is -ve as we up; y is +ve).
Or as explained earlier, pot is higher at higher points.
Therefore, the pot at the given point P, 4300.728 m above earth's surface,
VP = 0 + 430.0728 kV =430.0728 kV.