Q. 53.9( 9 Votes )
A dielectric slab A. increase
C. remain unchanged
D. become zero
We know that,
Force between the plates of the capacitor is given by
Q=charge on the capacitor
A=area of plates
Suppose charge Q and -Q are provided on plates of capacitor of area A.
Q is the test charge on the point charge
E is the electric field intensity.
Field due to charge Q on one plate is
Force on the plate with charge -Q will be
Considering magnitude, each plate applies a force of
On increasing a dielectric slab between the plates of the capacitor, the charge on the plates remains constant as the plates are isolated) .
Since, area of plates does not change, force between the plates remain constant.
Rate this question :
A parallel platePhysics - Exemplar
A parallel-plateHC Verma - Concepts of Physics Part 2
Two metal platesHC Verma - Concepts of Physics Part 2
The dielectric coHC Verma - Concepts of Physics Part 2
When a dielectricHC Verma - Concepts of Physics Part 2