Home
Class 12
PHYSICS
Find the surface density of electric ch...

Find the surface density of electric charge at a place on the earth's surface, where the rate of fall of potential is 2.5 V.

A

`2.0xx10^(-9)C//m^(2)`

B

`2.21xx10^(-9)C//m^(2)`

C

`3.36xx10^(-9)C//m^(2)`

D

`3.5xx10^(-9)C//m^(2)`

Text Solution

AI Generated Solution

Promotional Banner

Topper's Solved these Questions

  • GAUSS LAW AND ELECTRIC POTENTIAL THEORY

    BITSAT GUIDE|Exercise BITSTAT ARCHIVES|4 Videos
  • FLUID MECHANICS

    BITSAT GUIDE|Exercise BITSAT Achives|15 Videos
  • MAGNETIC FIELD

    BITSAT GUIDE|Exercise All Questions|52 Videos

Similar Questions

Explore conceptually related problems

Calculate the surface density of change at a place on the earth's surface where the rate of fall of potential is 250 volts per metre. [Hint : rate of fall of potential =electric =E=(sigma)/(epsilon_(0)) by Gauss's theorm.]

The surface density of charge on the earth's surface is 2.65xx10^(-9)C//m^(2) . If the radius of the earth is 6400 km then the charge carried by the earth is

A body of mass m rises to a height h=R//5 from the earth's surface where R is earth's radius. If g is acceleration due to gravity at the earth's surface, the increase in potential energy is

A spherical charged conductor has sigma as the surface density of charge. The electric field on its surface is E. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere -