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noppawit
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Earth's lower atmosphere contains negative and positive ions that are produced by radioactive elements in the soil and cosmic rays from space. In a certain region, the atmospheric electric field strength is 120V/m and the field is directed vertically down. This field causes singly charged positive ions, at a density of 620cm-3, to drift downward and singly charged negative ions, at a density of 550 cm-3, to drift upward. The measured conductivity of the air in the region is 2.70*10-14. Calculate a) the magnitude of the current density and b) the ion drift speed, assumed to be the same for positive and negative ions.
E=120V/m; n+ = 620cm-3=620*106m-3; n- = 550cm-3 = 550*106m-3
σ=2.7*10-14 (Ω*m)-1
σ = 1/ρ; ρ=E/J then 1/ρ = J/E
2.7*10-14 = J/120
J=2.7*120*10=3.24*10-12 A/m2 ===== a)
J=nevd = (n+ - n-)*e*vd
3.24*10-12 = (620-550)*10-6*1.602*10-19*vd
vd=0.289m/s
Am I correct? Is there anything wrong in my solution. Please tell me.
Thank you.
E=120V/m; n+ = 620cm-3=620*106m-3; n- = 550cm-3 = 550*106m-3
σ=2.7*10-14 (Ω*m)-1
σ = 1/ρ; ρ=E/J then 1/ρ = J/E
2.7*10-14 = J/120
J=2.7*120*10=3.24*10-12 A/m2 ===== a)
J=nevd = (n+ - n-)*e*vd
3.24*10-12 = (620-550)*10-6*1.602*10-19*vd
vd=0.289m/s
Am I correct? Is there anything wrong in my solution. Please tell me.
Thank you.