Q. 144.2( 25 Votes )

The earth’s surface has a negative surface charge density of 10-9 C m–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 neutralize 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.)

Answer :

Surface Charge density of earth d = 10-9 Coulomb per meter square

Current over the entire globe = 1800 Ampere

Radius of earth r = 6.37 × 106 meter

Surface area of earth A = 4 × π × radius × radius

A = 4 × π × 6.37 × 106 × 6.37 × 106

A = 5.09 × 1014 m2

Charge on the earth surface q = d × A

q = 10-9 × 5.09 × 1014 m2

q = 5.09 × 105 Coulomb

Let the time taken to neutralize earth surface = t


t = 5.09 × 105C/1800 A

t = 282.78 seconds.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Ohm's Law - Full Concept in one Go59 mins
Know about Internal Resistance & EMF42 mins
Interactive Quiz on Ohm's law & Drift + Thermal Velocity42 mins
Interactive Quiz on Current Electricity46 mins
An Elaborative lecture on basics of Current Electricity53 mins
Quick Revision of Current Electricity68 mins
Quiz | Wheatstone Bridge51 mins
Quiz | Drift Velocity & Thermal Velocity33 mins
Drift Velocity & Thermal Velocity34 mins
Kirchoff's Law for VoltageFREE Class
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses