In an isolted, charged, parllel-plate air capacitor, the charge per unit area on each plate has a magnitude of `sigma`. A dielectric slab having the dielectric constant K is now introduced between the plates. The induced charge per unit area on the surface of the dielectric will have magnitude.
A
`(sigma)/(K)`
B
`sigma(K-1)`
C
`sigma(1-(1)/(K))`
D
`(sigma)/(K+1)`
Text Solution
Verified by Experts
The correct Answer is:
C
Topper's Solved these Questions
PRACTICE WORKSHEET 3
D MUKHERJEE|Exercise Assertion- Reason Type|3 Videos
PRACTICE WORKSHEET 3
D MUKHERJEE|Exercise Linked- Comprehension Type|3 Videos
PRACTICE WORKSHEET 2
D MUKHERJEE|Exercise Linked- Comprehension Type|4 Videos
PROPERTIES OF MATTER,FLUIDS
D MUKHERJEE|Exercise Type 2|35 Videos
Similar Questions
Explore conceptually related problems
A parallel plate capacitor carries a harge Q. If a dielectric slab with dielectric constant K=2 is dipped between the plates, then
The separation between the plates of a parallel plate capacitor is d and the area of each plate is A. iff a dielectric slab of thickness x and dielectric constant K is introduced between the plates, then the capacitance will be
A parallel plate capacitor has a capacitance of "5pF" ,a plate area of 20.0mm^(2) and voltage across it is 10V.A dielectric slab with dielectric constant "5" is introduced in between the plates.Find the magnitude of the induced surface charge on the slab.
A parallel- plate capacitor is charged from a cell and then isolated from it. A dielectric slab of dielectric constant K is now introduced in the region between the plated, filling half of it. The electric intensity in the diaelectric is E_(1) and that in air is E_(2) .
A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K= (5)/(3) is inserted between the plates, the magnitude of the induced charge will be :
If a dielectric slab of thickness 5 mm and dielectric constant K=6 is introduced between the plates of a parallel plate air capacitor, with plate separation of 8 mm, then its capacitance is
Assertion : A parallel plate capacitor is connected across battery through a key. A dielectric slab dielectric constant K is introduced between the plates. The energy which is stored becomes K times. Reason : The surface density of charge on the plate remains constant or uncharged.
When a dielectric is introduced between the plates of a charged parallel plate capacitor, whichi one of the following will not change?
A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is Q and the magnitude of the induced charge on each surface of the dielectric is Q' .
D MUKHERJEE-PRACTICE WORKSHEET 3-Matrix-Matching Type