Test Index

NCERT Class XII Chapter
Current Electricity
Questions With Solutions

© examsnet.com
Question : 14
Total: 24
The earth’s surface has a negative surface charge density of 10–9cm–2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe).
(Radius of earth = 6.37 × 106 m).
Solution:  

Due to negative charge on earth, an electric field is into the earth surface due to which the positive ions of atmosphere are constantly pumped in and an equivalent current of 1800 A is established across the globe.
Let us first calculate total negative charge on earth
Q = σA = 10−9[4πRe2]
Q = 10−9[4π(6.37×106)2] or Q = 509.65 × 103 C
Now time to neutralize
I =
Q
t
t =
Q
I
 =
509.65×103
1800
 or t = 283.14 s
© examsnet.com
Go to Question:
Prev Question
Next Question