0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

If dielectric is inserted in charged capacitor (battery removed ), then quantity that remains constant is.
  1. Capacitance
  2. Charge
  3. Intensity
  4. Potential

A
Intensity
B
Charge
C
Potential
D
Capacitance
Solution
Verified by Toppr

Variation of different variables (Q,C,V,E and U) of parallel plate capacitor when dielectric (K) is introduced when battery is removed is
C=KC E=E/K
Q=Q U=U/K
V=V/K

Was this answer helpful?
2
Similar Questions
Q1
If dielectric is inserted in charged capacitor (battery removed ), then quantity that remains constant is.
View Solution
Q2
On removing dielectric slab from an isolated charged capacitor, the quantity that remains constant is
View Solution
Q3
A capacitor of capacitance C is connected to battery of emf V0. Without removing the battery, a dielectric of strength εr is inserted between the parallel plates of the capacitor C, then the charge on the capacitor is :
View Solution
Q4

A parallel plate capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E. If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become :


View Solution
Q5
A parallel plate capacitor is connected to a battery and a dielectric plate is inserted into the capacitor. Which quantity will increase?
View Solution