A parallel-plate capacitor having plate area 100 cm 2 and separation 1 . 0 mm holds a charge of 0 . 12 μ C μC when connected to a 120 V battery. Find the dielectric constant of the material filling the gap.

Given that plate area, A = 100 cm 2 , Separation between plates, d = 1 . 00 mm Relation between potential difference and charge is, q = C × V Capacitance, C = q V = 0 .12 μC 120 V = 10 - 9 F If k is the dielectric constant, then expression for capacitance, C = k ε 0 A d C = k × 8 .85 × 10 - 12 × 100 × 10 - 4 1 .0 × 10 - 3 = 10 - 9 F After simplification, ∴ k = 11 . 3

A parallel-plate capacitor consists of 26 metal strips, each of 3 cm × 4 cm , separated by mica sheets of dielectric constant 6 and uniform thickness 0 . 2 mm . Find the capacitance.

Given that, Number of metal strip n = 26 Each of cross- sectional area, A = 3 cm × 4 cm Dielectric constant of the paper, k = 6 Thickness of each parallel plate capacitor, d = 0 .2 mm = 0 . 2 × 10 - 3 m Arrangement of n metal plates separated by dielectric acts as a parallel combination of n - 1 capacitors. Capacitance of n - 1 capacitors is ∴ C = n - 1 k ε 0 A d = 25 × 6 × 8 .85 × 10 - 12 × 3 × 4 × 10 - 4 0 .2 × 10 - 3 = 7 .97 × 10 - 9 F

EASY

12th CBSE

IMPORTANT

Earn 100

Two capacitors are in parallel and the energy stored is $45 J$ , when the combination is raised to the potential of $3000 V$ . With the same two capacitors in series, the energy stored is $4.05 J$ for the same potential. What are their individual capacitances?

Important Questions on Electrostatic Potential and Capacitance

MEDIUM

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 2 - Electrostatic Potential and Capacitance>Problems For Practice>Q 10

Find the ratio of the potential differences that must be applied across the parallel and the series combination of two capacitors

C_{1}

and

C_{2}

with their capacitances in the ratio $1 : 2$ so that the energy stored in the two cases, becomes the same.

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 2 - Electrostatic Potential and Capacitance>Problems For Practice>Q 1

A parallel-plate capacitor having plate area

100 {cm}^{2}

and separation

1.0 mm

holds a charge of

U = \frac{1}{2} (\frac{C_{1} C_{2}}{C_{1} + C_{2}}) V^{2}

μC when connected to a

120 V

battery. Find the dielectric constant of the material filling the gap.

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 2 - Electrostatic Potential and Capacitance>Problems For Practice>Q 2

Find the length of the paper used in a capacitor of capacitance

2 μ F

, if the dielectric constant of the paper is

2.5

and its width and thickness are

50 mm

and

0.05 mm

, respectively.

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 2 - Electrostatic Potential and Capacitance>Problems For Practice>Q 3

A parallel-plate capacitor consists of

26

metal strips, each of

3 cm \times 4 cm

, separated by mica sheets of dielectric constant 6 and uniform thickness

0.2 mm

. Find the capacitance.

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 2 - Electrostatic Potential and Capacitance>Problems For Practice>Q 4

A parallel-plate capacitor of capacity

0.5 μ F

is to be constructed using paper sheets of thickness

0.04 mm

as dielectric. Find how many circular metal foils of diameter

0.1 m

will have to be used. Take the dielectric constant of paper used as

4

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 2 - Electrostatic Potential and Capacitance>Problems For Practice>Q 5

When a slab of insulating material

4 mm

thick is introduced between the plates of a parallel plate capacitor, it is found that the distance between the plates has to be increased by

3.2 mm

to restore the capacitance to the original value. Calculate the dielectric constant of the material.

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 2 - Electrostatic Potential and Capacitance>Problems For Practice>Q 6

The two plates of a parallel plate capacitor are

4 mm

apart. A slab of dielectric constant

3

and thickness

3 mm

is introduced between the plates with its faces parallel to them. The distance between the plates is so adjusted that the capacitance of the capacitor becomes

\frac{2}{3}

rd of its original value. What is the new distance between the plates?

MEDIUM

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 2 - Electrostatic Potential and Capacitance>Problems For Practice>Q 7

The distance between the parallel plates of a Charged capacitor is

5 cm

and the intensity of the electric field is

300 V {cm}^{- 1}

. A slab of dielectric constant 5 and thickness,

1 cm

is inserted parallel to the plates. Determine the potential difference between the plates, before and after the slab is inserted?

Two capacitors are in parallel and the energy stored is 45 J, when the combination is raised to the potential of 3000 V. With the same two capacitors in series, the energy stored is 4.05 J for the same potential. What are their individual capacitances?

Important Questions on Electrostatic Potential and Capacitance

Learn and Prepare for any exam you want !

Two capacitors are in parallel and the energy stored is $45 J$ , when the combination is raised to the potential of $3000 V$ . With the same two capacitors in series, the energy stored is $4.05 J$ for the same potential. What are their individual capacitances?