An infinite plane sheet of charge density `10^(-8) Cm^(-2)` is held in air. In this situation how far apart are two equipotenitial surfaces, whose p.d is 5 V ?

Answer

8.85 mm

Explanation

Electric field intensity of an infinite of an infinite plane sheet of charge, `E = (sigma)/(2 in_(0))`
Let `Delta` r be the surface between two equipotential surface having potential difference `Delta V`,
then `E = (Delta V)/(Delta r)`
`(sigma)/(2 in_(0)) = (Delta V)/(Delta r) implies Delta r = (2in_(0) Delta V)/(sigma)`
`= (2xx8.85xx10^(-12)xx5)/(10^(-8)) = 8.85xx10^(-3) m`
= 8.85 mm