Solve

Guides

0

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

Question

A capacitor of capacitance C is connected to battery of emf $V_{0}$ . Without removing the battery, a dielectric of strength $ε_{r}$ is inserted between the parallel plates of the capacitor C, then the charge on the capacitor is :
$C V_{0}$
$ε_{r} C V_{0}$
$\frac{C V_{0}}{ε_{r}}$
none of these

A

ε_{r} C V_{0}

B

none of these

C

C V_{0}

D

\frac{C V_{0}}{ε_{r}}

Open in App

Solution

Verified by Toppr

As the battery is not removed so the potential remains constant and it is equal to $V_{0}$ .
Due to insert the dielectric the capacitance $C$ will become $C^{'} = ε_{r} C$
Thus the charge on the capacitor $Q = C^{'} V_{0} = ε_{r} C V_{0}$

Was this answer helpful?

2

Similar Questions

Q1

A capacitor of capacitance C is connected to battery of emf

V_{0}

. Without removing the battery, a dielectric of strength

ε_{r}

is inserted between the parallel plates of the capacitor C, then the charge on the capacitor is :

View Solution

Q2

A capacitor of capacitance ‘C’ is connected with a battery of emf ε as shown. After full charging a dielectric of same size of capacitor and dielectric constant k is inserted then choose correct statements. (capacitor is always connected to battery)

View Solution

Q3

A capacitor of capacitance

5 μ F

is connected to a battery of emf

4 V .

Now this charged capacitor is disconnected with the battery and a dielectric of dielectric constant

2

is inserted between the plates. Find the magnitude of the charge induced on each surface of dielectric

View Solution

Q4

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 = \frac{5}{3}

is inserted between the plates, the magnitude of induced charge will be:

View Solution

Q5

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