Capacitors and Dielectrics HRK Physics: Complete Guide to Parallel Plate, Cylindrical & Spherical Capacitors

Chapter 31: Capacitors and Dielectrics

Complete Physics Notes Based on Halliday, Resnick and Krane - B.Sc. Physics Edition 2015-16

Capacitors Dielectrics Energy Storage Electromagnetism Reading Time: 30 min

📋 Table of Contents

• 1. Introduction to Capacitors
• 2. Capacitance Definition
• 3. Parallel Plate Capacitor
• 4. Cylindrical Capacitor
• 5. Spherical Capacitor
• 6. Isolated Sphere Capacitance
• 7. Energy Stored in Capacitors
• 8. Energy Density in Electric Fields
• 9. Capacitance with Dielectrics
• 10. Dielectric Materials
• 11. Gauss's Law in Dielectrics
• 12. Solved Problems
• Frequently Asked Questions

Introduction to Capacitors

⚡ What is a Capacitor?

A capacitor is a device that stores energy in an electrostatic field. It consists of two conductors separated by a small distance, with either vacuum or a dielectric medium between them.

🔬 Applications of Capacitors

• Energy Storage: Flash bulbs require short bursts of energy that exceed what batteries can provide. Capacitors draw energy slowly from batteries and release it rapidly.
• Laser Pulses: Large capacitors provide intense laser pulses for thermonuclear fusion experiments, reaching power levels of about \(10^{14}\) W for durations of \(10^{-9}\) s.
• Electric Field Production: Parallel plate capacitors create electric fields to deflect beams of charged particles.
• Electronic Circuits: Capacitors are fundamental components in radio and TV signal transmission and reception, and in electromagnetic oscillators.

Capacitance Definition

📏 Definition of Capacitance

When plates of a capacitor are connected to a battery of emf V, charge q is stored. This stored charge is directly proportional to the potential difference applied between the plates:

\[ q \propto V \]

↓

\[ q = CV \]

Here C is the constant of proportionality called capacitance. The SI unit of capacitance is the farad (F).

🎯 Definition of Farad

"If one coulomb of charge given to the plates produces a potential difference of one volt, then the capacitance of the capacitor is one farad."

Problem 1: Electrometer Minimum Charge Measurement

Question: An electrometer is a device used to measure static charge. Unknown charge is placed on the plates of a capacitor and the potential difference is measured. What minimum charge can be measured by an electrometer with a capacitance of 50 pF and a voltage sensitivity of 0.15 V?

Solution:

Given: \( C = 50 \, pF = 50 \times 10^{-12} F \), \( V = 0.15 \, V \)

\[ q = CV \]

\[ = 50 \times 10^{-12} \times 0.15 \]

\[ = 7.5 \times 10^{-12} \, C \]

Sample Problem 1: Excess Electrons on VRAM Chip

Question: A storage capacitor on a random access memory (VRAM) chip has a capacitance of 55 fF. If it is charged to 5.3 V, how many excess electrons are there on its negative plate?

Solution:

Given: \( C = 55 \, fF = 55 \times 10^{-15} F \), \( V = 5.3 \, V \)

\[ q = CV = 55 \times 10^{-15} \times 5.3 \]

\[ = 291 \times 10^{-15} \, C \]

Number of electrons \( n = \frac{q}{e} = \frac{291 \times 10^{-15}}{1.6 \times 10^{-19}} \]

\[ = 181.8 \times 10^4 = 1.818 \times 10^6 \, \text{electrons} \]

Parallel Plate Capacitor

🔋 Structure of Parallel Plate Capacitor

A parallel plate capacitor consists of two parallel conducting plates separated by a distance d, with area A for each plate. The plates carry equal and opposite charges.

31.1.1 Electric Field between Plates

Gaussian Surface Selection

Consider a box-shaped Gaussian surface that encloses part of one plate. The electric field inside a conductor is zero, and between plates it's uniform.

Apply Gauss's Law

\[ \oint \vec{E} \cdot d\vec{A} = \frac{q}{\epsilon_0} \]

\[ \Rightarrow EA = \frac{q}{\epsilon_0} \]

\[ \Rightarrow E = \frac{q}{A\epsilon_0} \]

\[ \Rightarrow E = \frac{\sigma}{\epsilon_0} \quad \text{where } \sigma = \frac{q}{A} \]

31.1.2 Potential Difference between Plates

Potential Difference Formula

\[ V_f - V_i = -\int_i^f \vec{E} \cdot d\vec{s} \]

For Parallel Plates

\[ V_f - V_i = -E \int_0^d ds \]

\[ = -E d \]

Final Expression

If initial plate is positively charged and final plate is negatively charged:

\[ V = E d \]

31.1.3 Capacitance of Parallel Plate Capacitor

Start with Electric Field

\[ E = \frac{q}{A\epsilon_0} \quad \text{(1)} \]

Relationship with Potential

\[ V = E d \Rightarrow E = \frac{V}{d} \quad \text{(2)} \]

Equate and Solve

\[ \frac{V}{d} = \frac{q}{A\epsilon_0} \]

\[ \Rightarrow \frac{q}{V} = \frac{A\epsilon_0}{d} \]

\[ \Rightarrow C = \frac{A\epsilon_0}{d} \]

With Dielectric Medium

\[ C = \frac{A\epsilon_0 \kappa_e}{d} \]

where \( \kappa_e \) is the dielectric constant.

Sample Problem 2: Plate Area for 1F Capacitor

Question: The plates of parallel plate capacitor are separated by a distance d=1mm. What must be the plate area if the capacitance is to be 1 F?

Solution:

Given: \( d = 1 \, mm = 1 \times 10^{-3} \, m \), \( C = 1 \, F \)

\[ C = \frac{A\epsilon_0}{d} \]

\[ \Rightarrow A = \frac{Cd}{\epsilon_0} = \frac{1 \times 1 \times 10^{-3}}{8.85 \times 10^{-12}} \]

\[ = 1.1 \times 10^8 \, m^2 \]

Problem 4: Circular Parallel Plate Capacitor

Question: A parallel plate capacitor has circular plates of 8.22 cm radius and 1.31 mm separation. (a) Calculate the capacitance. (b) What charge will appear on the plates if a potential difference of 116 V is applied?

Solution:

Given: \( r = 8.22 \, cm = 8.22 \times 10^{-2} \, m \), \( d = 1.31 \, mm = 1.31 \times 10^{-3} \, m \)

(a) Capacitance:

\[ C = \frac{A\epsilon_0}{d} = \frac{\pi r^2 \epsilon_0}{d} \]

\[ = \frac{3.14 \times (8.22 \times 10^{-2})^2 \times 8.85 \times 10^{-12}}{1.31 \times 10^{-3}} \]

\[ = 1433 \times 10^{-14} \, F = 14.33 \, pF \]

(b) Charge when \( V = 116 \, V \):

\[ q = CV = 14.33 \times 10^{-12} \times 116 \]

\[ = 1662 \times 10^{-12} \, C = 1.662 \, nC \]

Problem 43: Dielectric Constant of Wax

Question: An air filled parallel plate capacitor has capacitance of 1.32 pF. The separation of the plate is doubled and wax inserted between them. The new capacitance is 2.57 pF. What is dielectric constant of wax?

Solution:

Given: \( C_0 = 1.32 \, pF \), \( C = 2.57 \, pF \)

\[ C_0 = \frac{A\epsilon_0}{d} \]

\[ C = \frac{A\epsilon_0 \kappa_e}{2d} = \left( \frac{A\epsilon_0}{d} \right) \frac{\kappa_e}{2} = C_0 \frac{\kappa_e}{2} \]

\[ \Rightarrow \kappa_e = \frac{2C}{C_0} = \frac{2 \times 2.57 \times 10^{-12}}{1.32 \times 10^{-12}} \]

\[ = 3.89 \]

Cylindrical Capacitor

🔋 Structure of Cylindrical Capacitor

A cylindrical capacitor consists of two coaxial cylinders of radii 'a' and 'b', with length L. The inner cylinder is positively charged and the outer cylinder is negatively charged.

Electric Field Between Cylinders

Consider a cylindrical Gaussian surface of radius r (a < r < b):

\[ \oint \vec{E} \cdot d\vec{A} = \frac{q}{\epsilon_0} \]

\[ \Rightarrow E(2\pi rL) = \frac{q}{\epsilon_0} \]

\[ \Rightarrow E = \frac{q}{2\pi \epsilon_0 rL} \]

Potential Difference

\[ V = \int_a^b E \, dr = \int_a^b \frac{q}{2\pi \epsilon_0 rL} \, dr \]

\[ = \frac{q}{2\pi \epsilon_0 L} \int_a^b \frac{dr}{r} \]

\[ = \frac{q}{2\pi \epsilon_0 L} [\ln b - \ln a] \]

\[ = \frac{q}{2\pi \epsilon_0 L} \ln \left( \frac{b}{a} \right) \]

Capacitance

\[ C = \frac{q}{V} = \frac{2\pi \epsilon_0 L}{\ln \left( \frac{b}{a} \right)} \]

Problem 5: Vacuum Tube Diode Capacitance

Question: The plate and cathode of a vacuum tube diode are in the form of two concentric cylinders with the cathode as central cylinder. The cathode diameter is 1.62 mm and the plate diameter 18.3 mm with both elements having a length of 2.38 cm. Calculate the capacitance of the diode.

Solution:

Given: \( a = 1.62 \, mm = 1.62 \times 10^{-3} \, m \), \( b = 18.3 \, mm = 18.3 \times 10^{-3} \, m \), \( L = 2.38 \, cm = 2.38 \times 10^{-2} \, m \)

\[ C = \frac{2\pi \epsilon_0 L}{\ln \left( \frac{b}{a} \right)} \]

\[ = \frac{2 \times 3.14 \times 8.85 \times 10^{-12} \times 2.38 \times 10^{-2}}{\ln \left( \frac{18.3}{1.62} \right)} \]

\[ = \frac{132.3 \times 10^{-14}}{\ln (11.3)} \]

\[ = \frac{132.3 \times 10^{-14}}{2.42} \]

\[ = 54.6 \times 10^{-14} \, F = 0.546 \, pF \]

Spherical Capacitor

🔋 Structure of Spherical Capacitor

A spherical capacitor consists of two concentric spherical shells of radii 'a' and 'b' (b > a). The inner sphere is positively charged and the outer sphere is negatively charged.

Electric Field Between Shells

Consider a spherical Gaussian surface of radius r (a < r < b):

\[ \oint \vec{E} \cdot d\vec{A} = \frac{q}{\epsilon_0} \]

\[ \Rightarrow E(4\pi r^2) = \frac{q}{\epsilon_0} \]

\[ \Rightarrow E = \frac{q}{4\pi \epsilon_0 r^2} \]

Potential Difference

\[ V = \int_a^b E \, dr = \int_a^b \frac{q}{4\pi \epsilon_0 r^2} \, dr \]

\[ = \frac{q}{4\pi \epsilon_0} \int_a^b \frac{dr}{r^2} \]

\[ = \frac{q}{4\pi \epsilon_0} \left[ -\frac{1}{r} \right]_a^b \]

\[ = \frac{q}{4\pi \epsilon_0} \left( \frac{1}{a} - \frac{1}{b} \right) \]

Capacitance

\[ C = \frac{q}{V} = \frac{4\pi \epsilon_0}{\frac{1}{a} - \frac{1}{b}} \]

\[ = 4\pi \epsilon_0 \frac{ab}{b-a} \]

Problem 6: Earth-Ionosphere Capacitance

Question: What is the capacitance of Earth? Take it as an isolated sphere of radius 6370 km.

Solution:

Given: \( R = 6370 \, km = 6.37 \times 10^6 \, m \)

\[ C = 4\pi \epsilon_0 R \]

\[ = \frac{6.37 \times 10^6}{9 \times 10^9} \]

\[ = 7.08 \times 10^{-4} \, F = 708 \, \mu F \]

Isolated Sphere Capacitance

🌐 Capacitance of Isolated Sphere

An isolated sphere can be considered as a spherical capacitor where the outer sphere has infinite radius (b → ∞).

Start with Spherical Capacitor Formula

\[ C = 4\pi \epsilon_0 \frac{ab}{b-a} \]

Take Limit as b → ∞

\[ \lim_{b \to \infty} C = \lim_{b \to \infty} 4\pi \epsilon_0 \frac{ab}{b-a} \]

\[ = \lim_{b \to \infty} 4\pi \epsilon_0 \frac{a}{1 - \frac{a}{b}} \]

\[ = 4\pi \epsilon_0 a \]

📐 Final Expression

The capacitance of an isolated sphere of radius R is:

\[ C = 4\pi \epsilon_0 R \]

Energy Stored in Capacitors

⚡ Energy Storage in Capacitors

When a capacitor is charged, work is done to transfer charge from one plate to the other. This work is stored as potential energy in the electric field between the plates.

Work Done in Charging Process

At any instant during charging, the potential difference is v = q/C. The work done to add a small charge dq is:

\[ dW = v \, dq = \frac{q}{C} \, dq \]

Total Work

\[ W = \int_0^Q \frac{q}{C} \, dq \]

\[ = \frac{1}{C} \int_0^Q q \, dq \]

\[ = \frac{1}{C} \left[ \frac{q^2}{2} \right]_0^Q \]

\[ = \frac{Q^2}{2C} \]

Alternative Forms

Since Q = CV, we can write:

\[ U = \frac{Q^2}{2C} = \frac{1}{2} CV^2 = \frac{1}{2} QV \]

Sample Problem 7: Energy in Parallel Plate Capacitor

Question: A parallel plate capacitor has plates of area A and separation d. A potential difference V is applied between the plates. Calculate the stored energy.

Solution:

Using the energy formula:

\[ U = \frac{1}{2} CV^2 \]

\[ = \frac{1}{2} \left( \frac{A\epsilon_0}{d} \right) V^2 \]

\[ = \frac{A\epsilon_0 V^2}{2d} \]

Problem 8: Energy in Thundercloud

Question: A certain thundercloud has a potential difference of \( 1 \times 10^8 \, V \) relative to a tree. If during a lightning flash 50 C of charge is transferred, what is the change in energy of that charge?

Solution:

Given: \( V = 1 \times 10^8 \, V \), \( Q = 50 \, C \)

\[ U = \frac{1}{2} QV \]

\[ = \frac{1}{2} \times 50 \times 1 \times 10^8 \]

\[ = 2.5 \times 10^9 \, J \]

Energy Density in Electric Fields

📊 Energy Density Concept

The energy stored in a capacitor is not localized on the plates but is distributed throughout the electric field between the plates. The energy density u is the energy per unit volume.

Start with Parallel Plate Capacitor Energy

\[ U = \frac{1}{2} CV^2 \]

Substitute C and V

\[ U = \frac{1}{2} \left( \frac{A\epsilon_0}{d} \right) (Ed)^2 \]

\[ = \frac{1}{2} \frac{A\epsilon_0}{d} E^2 d^2 \]

\[ = \frac{1}{2} \epsilon_0 E^2 (Ad) \]

Energy Density

Since Ad is the volume between plates:

\[ u = \frac{U}{\text{Volume}} = \frac{1}{2} \epsilon_0 E^2 \]

🔍 General Result

The energy density in any electric field (not just in capacitors) is:

\[ u = \frac{1}{2} \epsilon_0 E^2 \]

This result is valid for all electric fields, regardless of their source.

Problem 9: Energy Density in Thundercloud

Question: The average breakdown electric field in air is \( 3 \times 10^6 \, V/m \). What is the energy density in this field?

Solution:

Given: \( E = 3 \times 10^6 \, V/m \)

\[ u = \frac{1}{2} \epsilon_0 E^2 \]

\[ = \frac{1}{2} \times 8.85 \times 10^{-12} \times (3 \times 10^6)^2 \]

\[ = \frac{1}{2} \times 8.85 \times 10^{-12} \times 9 \times 10^{12} \]

\[ = 39.8 \, J/m^3 \]

Capacitance with Dielectrics

🔬 Dielectric Materials

A dielectric is an insulating material that, when placed between the plates of a capacitor, increases its capacitance. The dielectric constant \( \kappa_e \) is defined as:

\[ \kappa_e = \frac{C}{C_0} \]

where \( C_0 \) is the capacitance with vacuum between plates and C is the capacitance with dielectric.

Effect on Electric Field

The electric field inside a dielectric is reduced by a factor of \( \kappa_e \):

\[ E = \frac{E_0}{\kappa_e} \]

Effect on Potential

For the same charge, the potential difference decreases:

\[ V = \frac{V_0}{\kappa_e} \]

Effect on Capacitance

The capacitance increases by a factor of \( \kappa_e \):

\[ C = \kappa_e C_0 \]

Problem 10: Paper Dielectric Capacitor

Question: A parallel plate capacitor has plates of area \( 2 \, m^2 \) separated by 1 cm. What is its capacitance if the space between plates is filled with paper of dielectric constant 3.7?

Solution:

Given: \( A = 2 \, m^2 \), \( d = 1 \, cm = 0.01 \, m \), \( \kappa_e = 3.7 \)

\[ C_0 = \frac{A\epsilon_0}{d} = \frac{2 \times 8.85 \times 10^{-12}}{0.01} \]

\[ = 1.77 \times 10^{-9} \, F \]

\[ C = \kappa_e C_0 = 3.7 \times 1.77 \times 10^{-9} \]

\[ = 6.55 \times 10^{-9} \, F = 6.55 \, nF \]

Dielectric Materials

🧲 Types of Dielectrics

Dielectrics can be classified based on their molecular structure and behavior in electric fields.

Nonpolar Dielectrics

Molecules have no permanent dipole moment. Examples: \( O_2 \), \( N_2 \), \( CH_4 \). In an external field, induced dipoles are created.

Polar Dielectrics

Molecules have permanent dipole moments. Examples: \( H_2O \), \( NH_3 \). In an external field, dipoles tend to align with the field.

Polarization of Dielectrics

When a dielectric is placed in an electric field, it becomes polarized. This means that electric dipoles are induced or aligned within the material.

Induced Surface Charges

Polarization creates induced surface charges on the dielectric. These charges create an electric field that opposes the external field.

Net Electric Field

\[ E = E_0 - E_{\text{induced}} \]

\[ = \frac{E_0}{\kappa_e} \]

📊 Common Dielectric Constants

• Vacuum: \( \kappa_e = 1 \)
• Air (1 atm): \( \kappa_e = 1.00054 \)
• Paper: \( \kappa_e = 3.7 \)
• Glass: \( \kappa_e = 5-10 \)
• Water: \( \kappa_e = 80 \)
• Titanium dioxide: \( \kappa_e = 100 \)

Gauss's Law in Dielectrics

📐 Modified Gauss's Law

When dielectrics are present, Gauss's law must be modified to account for both free charges and bound (induced) charges.

Original Gauss's Law

\[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{free}} + q_{\text{bound}}}{\epsilon_0} \]

Introduce Electric Displacement

Define the electric displacement vector \( \vec{D} = \epsilon_0 \vec{E} + \vec{P} \), where \( \vec{P} \) is the polarization vector.

Gauss's Law in Dielectrics

\[ \oint \vec{D} \cdot d\vec{A} = q_{\text{free}} \]

For Linear Dielectrics

For linear dielectrics, \( \vec{P} = \epsilon_0 \chi_e \vec{E} \), where \( \chi_e \) is the electric susceptibility.

\[ \vec{D} = \epsilon_0 \vec{E} + \epsilon_0 \chi_e \vec{E} = \epsilon_0 (1 + \chi_e) \vec{E} \]

\[ = \epsilon_0 \kappa_e \vec{E} \]

🔍 Key Points

• Gauss's law in dielectrics involves only free charges, not bound charges.
• The electric displacement \( \vec{D} \) depends only on free charges.
• The relationship between \( \vec{D} \) and \( \vec{E} \) depends on the dielectric properties.

Solved Problems

Problem 11: Cylindrical Capacitor with Dielectric

Question: A cylindrical capacitor has length L = 6.28 cm, inner radius a = 1 mm, outer radius b = 4 mm. The space between cylinders is filled with paper of dielectric constant 3.7. Calculate the capacitance.

Solution:

Given: \( L = 6.28 \, cm = 6.28 \times 10^{-2} \, m \), \( a = 1 \, mm = 1 \times 10^{-3} \, m \), \( b = 4 \, mm = 4 \times 10^{-3} \, m \), \( \kappa_e = 3.7 \)

\[ C_0 = \frac{2\pi \epsilon_0 L}{\ln(b/a)} \]

\[ = \frac{2 \times 3.14 \times 8.85 \times 10^{-12} \times 6.28 \times 10^{-2}}{\ln(4)} \]

\[ = \frac{34.9 \times 10^{-14}}{1.386} \]

\[ = 25.2 \times 10^{-14} \, F \]

\[ C = \kappa_e C_0 = 3.7 \times 25.2 \times 10^{-14} \]

\[ = 93.2 \times 10^{-14} \, F = 0.932 \, pF \]

Problem 12: Spherical Capacitor with Dielectric

Question: A spherical capacitor has inner radius a = 2 cm and outer radius b = 2.2 cm. The space between spheres is filled with oil of dielectric constant 4.5. Calculate the capacitance.

Solution:

Given: \( a = 2 \, cm = 2 \times 10^{-2} \, m \), \( b = 2.2 \, cm = 2.2 \times 10^{-2} \, m \), \( \kappa_e = 4.5 \)

\[ C_0 = 4\pi \epsilon_0 \frac{ab}{b-a} \]

\[ = \frac{2 \times 10^{-2} \times 2.2 \times 10^{-2}}{9 \times 10^9 \times (2.2 - 2) \times 10^{-2}} \]

\[ = \frac{4.4 \times 10^{-4}}{9 \times 10^9 \times 0.2 \times 10^{-2}} \]

\[ = \frac{4.4 \times 10^{-4}}{1.8 \times 10^7} \]

\[ = 2.44 \times 10^{-11} \, F \]

\[ C = \kappa_e C_0 = 4.5 \times 2.44 \times 10^{-11} \]

\[ = 1.098 \times 10^{-10} \, F = 109.8 \, pF \]

Problem 13: Energy in Parallel Plate Capacitor

Question: A parallel plate capacitor has plates of area 0.12 \( m^2 \) and separation 1.2 cm. A potential difference of 120 V is applied. Calculate (a) the capacitance, (b) the charge on each plate, (c) the energy stored.

Solution:

Given: \( A = 0.12 \, m^2 \), \( d = 1.2 \, cm = 0.012 \, m \), \( V = 120 \, V \)

(a) Capacitance:

\[ C = \frac{A\epsilon_0}{d} = \frac{0.12 \times 8.85 \times 10^{-12}}{0.012} \]

\[ = 8.85 \times 10^{-11} \, F = 88.5 \, pF \]

(b) Charge:

\[ Q = CV = 8.85 \times 10^{-11} \times 120 \]

\[ = 1.062 \times 10^{-8} \, C = 10.62 \, nC \]

(c) Energy stored:

\[ U = \frac{1}{2} CV^2 = \frac{1}{2} \times 8.85 \times 10^{-11} \times (120)^2 \]

\[ = 6.37 \times 10^{-7} \, J \]

Frequently Asked Questions

❓ What happens to the capacitance when a dielectric is inserted between the plates of a capacitor?

The capacitance increases by a factor equal to the dielectric constant of the material. For example, if a dielectric with \( \kappa_e = 3 \) is inserted, the capacitance becomes three times its original value.

❓ Why does the energy stored in a capacitor decrease when a dielectric is inserted while the capacitor remains connected to a battery?

When the capacitor remains connected to a battery, the potential difference remains constant. Inserting a dielectric increases the capacitance, which means more charge flows from the battery to the capacitor. However, the energy stored is given by \( U = \frac{1}{2} CV^2 \), so with constant V and increased C, the energy stored actually increases, not decreases.

❓ What is the difference between polar and nonpolar dielectrics?

Polar dielectrics have molecules with permanent electric dipole moments (e.g., water), while nonpolar dielectrics have molecules with no permanent dipole moments (e.g., oxygen). In an external electric field, polar dielectrics experience alignment of existing dipoles, while nonpolar dielectrics develop induced dipoles.

❓ Can the capacitance of a capacitor be negative?

No, capacitance is always a positive quantity. It represents the ability of a system to store charge per unit voltage, and both charge and voltage are positive quantities in this context.

❓ Why is the energy density in an electric field proportional to the square of the electric field?

The energy density \( u = \frac{1}{2} \epsilon_0 E^2 \) comes from the work done in establishing the electric field. Since the electric force is proportional to E, and work is force times distance, the energy stored ends up being proportional to \( E^2 \).

📚 Continue Your Physics Journey

Mastering Capacitors and Dielectrics is fundamental to understanding electromagnetism and circuit theory. These comprehensive notes based on Halliday, Resnick and Krane provide a solid foundation for further studies in physics including current electricity, alternating currents, and electromagnetic waves.

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These comprehensive notes are designed to help B.Sc. Physics students understand fundamental concepts of Capacitors and Dielectrics based on Halliday, Resnick and Krane

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