A fully charged parallel-plate capacitor remains connected to a battery while you slide a dielectric between the plates. Then the capacitance :
A
Increase
B
Stay the same
C
Decrease
D
Can not say.
Open in App
Solution
Verified by Toppr
Correct option is A. Increase Because $$C=k\epsilon \,A/d$$ and the dielectric constant $$k$$ increases. So C also increases.
Was this answer helpful?
0
Similar Questions
Q1
A fully charged parallel-plate capacitor remains connected to a battery while you slide a dielectric between the plates. Then the capacitance :
View Solution
Q2
A fully charged parallel-plate capacitor remains connected to a battery while you slide a dielectric between the plates. Do the following quantities the energy stored in the capacitor
View Solution
Q3
A parallel plate capacitor of capacitance 12 pF is connected to 10V battery, when it is fully charged battery is disconnected and dielectric of constant 6 is introduced between the plates then find energy dissipated during insertion of dielectric.
View Solution
Q4
A parallel plate capacitor (without dielectric) is charged by a battery and kept connected to the battery. A dielectric slab of dielectric constant ‘k′ is inserted between the plates fully occupying the space between the plates. The energy density of electric field between the plates will:
View Solution
Q5
The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is-