CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

New Capacitance in Dielectric

The capacitan...

Question

The capacitance of parallel-plate capacitor is $4 μ F$ . If a dielectric material of dielectric constant 16 is placed between the plates then the new capacitance will be :

A

1 / 64 μ F

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

0.25 μ F

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

64 μ F

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

40 μ F

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $64 μ F$
If the capacitance of the parallel plate capacitor without dielectric is $C$ , the capacitance of it with dielectric will be $C^{'} = k C = 16 \times 4 = 64 μ F$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Half of the space between the plates of parallel plate capacitor is filled with a medium of dielectric constant

K

parallel to the plates. If initially the capacitance is

C

, then the new capacitance will be

Q. A parallel-plate air capacitor has a capacitance of

4 μ F

what be its new capacitance if a slab of dielectric constant 5 is introduced filling the entire space between the two plates?

Q. For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant

K

is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is

\frac{3}{4} d,

^{'} d^{'}

is the separation between the plates of the parallel plate capacitor. The new capacitance

(C^{'})

in terms of the original capacitance

(C_{0})

is given by the following relation:

Q. Two parallel-plate capacitors, each of capacitance

40 μ F

, are connected in series. The space between the plates of one capacitor is filled with a dielectric of dielectric constant

K = 3

, then the equivalent capacitance of the combination is

Q. In a capacitor of capacitance

20 μ F

the distance between the plates is

2 m m .

If a dielectric slab of width

1 m m

and dielectric constant

2

is inserted between the plates, then the new capacitance will be:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Dielectrics

PHYSICS

Watch in App

explore_more_icon

Explore more

New Capacitance in Dielectric

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon