The plates of a parallel plate capacitor are charged upto 100 V. A 2 mm thich insulator sheet is inserted between the plates . Then to maintain the same potential difference the distance between the plates is increased by 1.6 mm. The dielectric constant of the insulator is

Options

A. 6
B. 8
C. 5
D. 4

Answer

C

Explanation

Potential of capacitor before inserting insulator .
`V= (sigma d)/(epsi_(0))`
where `sigma` = surface charge density d= distance between plates and `epsi_(0) ` = permittivity of vacuum.
Given `(sigmad)/(epsi_(0))= 100`
Now potential after inserting 2mm of insulator.
Let its dielectric constant is k.
`V_(1) = E_(1).2+ E_(2)(d-2)`
`V_(1)= (sigma)/(epsi_(0)k) 2 + (sigma)/(epsi_(0))(d-2) implies V_(1)= (sigma)/(epsi_(0)) ((2)/(k)+d-2)`
To maintain potential difference same distance is increased by 1.6 mm
So extra potential difference `V_(2)= (sigma)/(epsi_(0))(1.6)`
Clearly `V_(1)+V_(2)= V`
`(sigma)/(epsi_(0)) ((2)/(k)+d-2) + (sigma)/(epsi_(0))1.6 = (sigmad)/(epsi_(0))`
`implies (sigmad)/(epsi_(0))= (2sigma)/(epsi_(0)k)- (sigma)/(epsi_(0))(d-2)= (sigma)/(epsi_(0))1.6`
`implies (2sigma)/(epsi_(0))-(2sigma)/(epsi_(0)k)= (sigma)/(epsi_(0))1.6" " implies 2 -(2)/(k) = 1.6`
`implies " " 2k-2= 1.6 k`
`implies " " 0.4k= 2implies k =(2)/(0.4)`
`implies " " k=5`