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Chapter 2 Electrostatic Potential And Capacitance
Potential Difference And Electrostatic Potential
This chapter delves into the concepts of electrostatic potential energy and potential, building upon the understanding of conservative forces like Coulomb force. Just as gravitational force leads to gravitational potential energy, Coulomb force leads to electrostatic potential energy. This potential energy represents the work done by an external force to move a charge against the electrostatic field without acceleration. The potential energy difference between two points is path-independent, a characteristic of conservative forces. A convenient convention is to set the potential energy to zero at infinity, allowing for the definition of absolute potential energy at any point.
Figure 2.1 illustrates the process of moving a positive test charge against a repulsive force, highlighting the work done and its relation to potential energy.
Electrostatic Potential
Electrostatic potential ($V$) at a point in an electric field is defined as the work done per unit positive charge by an external force in bringing a test charge from infinity to that point, without acceleration. Mathematically, $V = W/q$, where $W$ is the work done and $q$ is the charge. This quantity is characteristic of the electric field itself and is independent of the test charge used for measurement. The unit of electrostatic potential is the Volt (V), where 1 V = 1 J/C. Like potential energy, the absolute value of potential is arbitrary; it is the potential difference that is physically significant.
Figure 2.2 emphasizes the path independence of work done in an electrostatic field. Figure 2.4 shows the variation of potential $V \propto 1/r$ and electric field $E \propto 1/r^2$ with distance $r$ from a point charge.
Potential Due To A Point Charge
The electrostatic potential at a distance $r$ from a point charge $Q$ placed at the origin is given by:
$V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$
This formula holds for both positive and negative values of $Q$. If $Q$ is positive, the potential is positive, meaning positive work is done by an external force to bring a positive charge from infinity. If $Q$ is negative, the potential is negative, indicating that the electrostatic force is attractive, and work is done by the field.
Example 2.1: Calculates the potential at a point due to a given charge and then uses this to find the work done in moving another charge from infinity to that point. It confirms that the work done is path-independent.
Potential Due To An Electric Dipole
An electric dipole consists of two equal and opposite charges, $q$ and $-q$, separated by a distance $2a$. The potential due to a dipole at a point $P$ at a distance $r$ from its center depends on both the distance $r$ and the angle $\theta$ between the position vector and the dipole moment vector $\vec{p}$. For distances large compared to the dipole size ($r \gg a$), the potential is given by:
$V(r, \theta) = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}$
where $p = q(2a)$ is the magnitude of the dipole moment. Key observations are:
- The potential depends on the angle $\theta$, indicating axial symmetry about the dipole moment.
- The potential falls off as $1/r^2$, which is faster than the $1/r$ dependence for a single point charge.
- On the dipole axis ($\theta = 0$ or $\pi$), $V = \pm \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$.
- In the equatorial plane ($\theta = \pi/2$), the potential is zero.
Example 2.2: Locates points on the line joining two charges where the electric potential is zero, demonstrating how potentials from individual charges add up.
Example 2.3: Explores the relationship between potential and field lines, analyzing potential differences, potential energy changes, and work done for positive and negative charges in different regions.
Potential Due To A System Of Charges
The electrostatic potential at any point due to a system of discrete charges is the algebraic sum of the potentials due to each individual charge at that point, according to the superposition principle:
$V = V_1 + V_2 + \ldots + V_n = \sum_{i=1}^{n} \frac{1}{4\pi\epsilon_0} \frac{q_i}{r_{iP}}$
where $q_i$ is the $i$-th charge and $r_{iP}$ is the distance from $q_i$ to the point $P$. For continuous charge distributions, this sum becomes an integral.
The potential outside a uniformly charged spherical shell is the same as that of a point charge located at its center, $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$ for $r \ge R$. Inside the shell ($r < R$), the potential is constant and equal to the potential at the surface, $V = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$, because the electric field inside is zero.
Equipotential Surfaces
An equipotential surface is a surface where the electrostatic potential is constant at all points. For a single point charge, these surfaces are concentric spheres centered on the charge. A key property is that the electric field at any point is perpendicular to the equipotential surface passing through that point. This is because no work is done in moving a charge along an equipotential surface.
- For a point charge: Equipotential surfaces are spheres centered on the charge.
- For a uniform electric field: Equipotential surfaces are planes perpendicular to the field lines.
- For a dipole or two like charges: Equipotential surfaces exhibit more complex shapes but maintain the property of being perpendicular to the electric field lines.
Relation Between Field And Potential
The electric field ($E$) and the potential ($V$) are related. The electric field points in the direction of the steepest decrease in potential. The magnitude of the electric field is equal to the rate of change of potential with distance in the direction of the field:
$E = -\frac{dV}{dl}$
where $dl$ is a small displacement along the direction of the electric field. This means electric field lines point from regions of higher potential to regions of lower potential.
Potential Energy Of A System Of Charges
The potential energy ($U$) of a system of charges is the work done by an external agency to assemble the charges from infinity to their given positions. For a system of two point charges $q_1$ and $q_2$ separated by a distance $r_{12}$, the potential energy is:
$U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}$
If the charges are like (both positive or both negative), the potential energy is positive, as work must be done against the repulsive force. If the charges are unlike, the potential energy is negative, as work is done by the attractive force when assembling them from infinity.
For a system of multiple charges, the total potential energy is the sum of the potential energies of all unique pairs of charges.
Example 2.4: Calculates the work required to assemble a system of four charges at the corners of a square and then the additional work to bring a charge to the center. It illustrates that the total energy depends on the final configuration, not the process of assembly.
Potential Energy In An External Field
Potential Energy Of A Single Charge
The potential energy of a single charge $q$ at a point with external potential $V(r)$ is given by:
$U(r) = qV(r)$
This assumes the potential $V(r)$ is due to sources external to the charge $q$, and that $q$ does not significantly alter the external field. The unit of energy commonly used in atomic and particle physics is the electron volt (eV), where $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$.
Potential Energy Of A System Of Two Charges In An External Field
For a system of two charges $q_1$ and $q_2$ at positions $\vec{r}_1$ and $\vec{r}_2$ in an external field, the potential energy is the sum of the potential energies of each charge in the external field and the mutual potential energy between the two charges:
$U = q_1 V(\vec{r}_1) + q_2 V(\vec{r}_2) + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}$
Example 2.5: Calculates the electrostatic energy of a system of two charges and the work required to separate them. It also considers the case where the system is placed in an external electric field.
Potential Energy Of A Dipole In An External Field
When an electric dipole with dipole moment $\vec{p}$ is placed in a uniform external electric field $\vec{E}$, it experiences a torque $\vec{\tau} = \vec{p} \times \vec{E}$. The potential energy $U$ of the dipole in the external field is given by:
$U(\theta) = -\vec{p} \cdot \vec{E} = -pE\cos\theta$
where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$. The potential energy is minimum when the dipole is aligned with the field ($\theta = 0$), and maximum when it is aligned antiparallel ($\theta = \pi$).
Example 2.6: Estimates the heat released by a substance when its dipoles align with a changed external electric field, relating it to the change in potential energy.
Electrostatics Of Conductors
Key properties of conductors in electrostatic situations include:
- Zero Electric Field Inside: The electrostatic field is zero everywhere inside a conductor in equilibrium.
- Normal Field at the Surface: The electric field at the surface of a charged conductor is always normal to the surface.
- No Excess Charge Inside: Any excess charge resides entirely on the surface of the conductor.
- Constant Potential Inside and on Surface: The electrostatic potential is constant throughout the volume of a conductor and has the same value on its surface.
- Electric Field at the Surface: The electric field just outside the surface of a conductor with surface charge density $\sigma$ is given by $E = \frac{\sigma}{\epsilon_0}$, directed normal to the surface.
- Electrostatic Shielding: A cavity within a conductor, even if charged externally, remains shielded from electric fields. All charges reside on the outer surface of such a conductor.
These properties arise from the free movement of charge carriers within the conductor, which rearrange themselves until equilibrium is reached.
Example 2.7: Explains phenomena like why a comb attracts paper, why aircraft tires are conductive, why vehicles carrying flammable materials have grounding ropes, and why a bird on a high-power line is safe, based on electrostatic principles.
Figure 2.17 illustrates the Gaussian surface used to derive the electric field at the surface of a conductor. Figure 2.18 and Figure 2.19 summarize key electrostatic properties of conductors and electrostatic shielding.
Dielectrics And Polarisation
Dielectrics are insulating materials. When placed in an external electric field, their molecules can become polarized. This polarization involves either the alignment of permanent molecular dipoles (in polar dielectrics) or the induction of dipole moments due to charge displacement within molecules (in non-polar dielectrics). The induced dipole moments create an internal electric field that opposes the external field, reducing the net field within the dielectric.
The extent of polarization is characterized by the dielectric susceptibility ($\chi_e$) and the permittivity of the medium ($\epsilon = \epsilon_0 K$), where $K$ is the dielectric constant. The dielectric constant ($K > 1$) represents the factor by which the capacitance increases when a dielectric material fills the space between the capacitor plates.
Figure 2.20 shows the difference in behavior between conductors and dielectrics in an external field. Figure 2.21 provides examples of polar and non-polar molecules. Figure 2.22 illustrates polarization in dielectrics, and Figure 2.23 shows how a polarized dielectric is equivalent to induced surface charges.
Capacitors And Capacitance
A capacitor is a device consisting of two conductors separated by an insulating material (dielectric). It is used to store electric charge and energy. The capacitance ($C$) of a capacitor is defined as the ratio of the magnitude of the charge ($Q$) on either conductor to the potential difference ($V$) between them:
$C = \frac{Q}{V}$
The capacitance is a measure of the capacitor's ability to store charge and depends only on the geometry (shape, size, separation of conductors) and the dielectric material between them. The SI unit of capacitance is the Farad (F), where $1 \text{ F} = 1 \text{ C/V}$. A Farad is a very large unit, and practical capacitors are typically measured in microfarads ($\mu$F), nanofarads (nF), or picofarads (pF).
Capacitors are essential components in electronic circuits for various applications, including filtering, tuning, and energy storage.
The Parallel Plate Capacitor
A parallel plate capacitor consists of two large, parallel conducting plates of area $A$ separated by a small distance $d$. With vacuum between the plates, the capacitance is given by:
$C_0 = \epsilon_0 \frac{A}{d}$
The electric field between the plates is uniform and localized, with magnitude $E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}$, where $\sigma$ is the surface charge density. The potential difference between the plates is $V = Ed = \frac{Qd}{\epsilon_0 A}$.
The large physical dimensions required for a capacitance of 1 Farad are highlighted, explaining why sub-multiples of Farad are commonly used.
Figure 2.25 illustrates the parallel plate capacitor and the electric field distribution.
Effect Of Dielectric On Capacitance
When a dielectric material is inserted between the plates of a capacitor, it reduces the electric field and hence the potential difference between the plates for a given charge. This leads to an increase in capacitance. If a dielectric material with dielectric constant $K$ completely fills the space between the plates, the new capacitance $C$ is related to the vacuum capacitance $C_0$ by:
$C = K C_0 = K \epsilon_0 \frac{A}{d}$
The dielectric constant $K$ is always greater than 1 for dielectric materials. This effect is crucial for increasing the charge storage capacity of capacitors.
Example 2.8: Calculates the change in capacitance when a dielectric slab of thickness $(3/4)d$ is inserted between the plates of a parallel plate capacitor.
Combination Of Capacitors
Capacitors In Series
When capacitors are connected in series, the charge on each capacitor is the same, and the total potential difference across the combination is the sum of the potential differences across individual capacitors. The effective capacitance $C$ for a series combination of capacitors $C_1, C_2, \ldots, C_n$ is given by:
$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}$
The equivalent capacitance in series is always less than the smallest individual capacitance.
Figure 2.26 and Figure 2.27 illustrate capacitors connected in series.
Capacitors In Parallel
When capacitors are connected in parallel, the potential difference across each capacitor is the same, and the total charge stored is the sum of the charges on individual capacitors. The effective capacitance $C$ for a parallel combination is:
$C = C_1 + C_2 + \ldots + C_n$
The equivalent capacitance in parallel is always greater than the largest individual capacitance.
Figure 2.28 shows capacitors connected in parallel.
Example 2.9: Calculates the equivalent capacitance of a network of capacitors connected in series and parallel, and determines the charge on each capacitor.
Energy Stored In A Capacitor
The electrostatic potential energy ($U$) stored in a capacitor with capacitance $C$, charge $Q$, and voltage $V$ can be expressed in three equivalent forms:
$U = \frac{1}{2} Q V = \frac{1}{2} C V^2 = \frac{1}{2} \frac{Q^2}{C}$
This energy is stored in the electric field between the capacitor plates. The energy density ($u$), which is the energy stored per unit volume, is given by:
$u = \frac{1}{2} \epsilon_0 E^2$
where $E$ is the electric field strength.
When capacitors are reconfigured (e.g., connecting charged capacitors to uncharged ones), energy can be lost as heat and radiation due to transient currents, even if charge is conserved.
Example 2.10: Calculates the electrostatic energy stored in a capacitor when charged by a battery and the energy stored after connecting two identical capacitors in parallel.
Exercises
Question 2.1. Two charges $5 \times 10^{–8}\text{ C}$ and $–3 \times 10^{–8}\text{ C}$ are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Answer:
Question 2.2. A regular hexagon of side 10 cm has a charge $5 \mu\text{C}$ at each of its vertices. Calculate the potential at the centre of the hexagon.
Answer:
Question 2.3. Two charges $2 \mu\text{C}$ and $–2 \mu\text{C}$ are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?
Answer:
Question 2.4. A spherical conductor of radius 12 cm has a charge of $1.6 \times 10^{–7}\text{C}$ distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?
Answer:
Question 2.5. A parallel plate capacitor with air between the plates has a capacitance of 8 pF ($1\text{pF} = 10^{–12}\text{ F}$). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Answer:
Question 2.6. Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
Answer:
Question 2.7. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Answer:
Question 2.8. In a parallel plate capacitor with air between the plates, each plate has an area of $6 \times 10^{–3}\text{ m}^2$ and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Answer:
Question 2.9. Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected.
Answer:
Question 2.10. A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Answer:
Question 2.11. A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Answer: