Chapter Notes

Introduction

In physics, some forces are known as conservative forces. When you do work against a conservative force, that work is stored as potential energy. Gravity and the force from a spring are great examples. If you lift a book, you do work against gravity, and that work is stored as gravitational potential energy. If you let go, the book falls, and the stored potential energy is converted into kinetic energy (the energy of motion).

The Coulomb force between stationary electric charges is also a conservative force. This is because, like gravity, it follows an inverse-square law with distance. This similarity allows us to define electrostatic potential energy for a charge in an electric field, just as we have gravitational potential energy for a mass in a gravitational field.

Electrostatic Potential Energy

Imagine an electric field E created by a charge Q at the origin. If we want to move a small positive test charge q from a point R to a point P against the repulsive force from Q, we must apply an external force.

There are two key conditions for this process:

1. The test charge q is so small that it doesn't affect the original charge configuration.
2. We move the charge q very slowly (with "infinitesimally slow constant speed"). This means the external force we apply, $\mathbf{F}_{\text{ext}}$, is exactly equal and opposite to the repulsive electric force, $\mathbf{F}_{\text{E}}$. So, $\mathbf{F}_{\text{ext}} = -\mathbf{F}_{\text{E}}$.

Under these conditions, the work done by the external force is stored as potential energy. The work done in moving the charge q from R to P is given by: $W_{RP} = \int_{R}^{P} \mathbf{F}_{ext} \cdot d\mathbf{r}$

This work increases the potential energy of the charge q. The change in potential energy, or the potential energy difference, is defined as the work done by the external force. $\Delta U = U_P - U_R = W_{RP}$

A fundamental property of conservative forces is that the work done is path-independent. It only depends on the starting point (R) and the ending point (P), not the path taken between them.

Defining a Zero Point for Potential Energy

The actual value of potential energy isn't physically significant; only the difference in potential energy matters. This gives us the freedom to choose a convenient reference point where the potential energy is zero. By convention, we choose infinity as the point of zero electrostatic potential energy ($U_{\infty} = 0$).

With this choice, the potential energy of a charge q at any point P is defined as the work done by an external force in bringing the charge q from infinity to that point P. $W_{\infty P} = U_P - U_{\infty} = U_P$ So, Potential Energy at a point P is: $U_P = W_{\infty P}$

Electrostatic Potential

The work done to move a charge q in an electric field is directly proportional to q. To create a quantity that describes the electric field itself, independent of the test charge being moved, we can calculate the work done per unit charge. This quantity is called electrostatic potential (V).

The potential difference between two points P and R is the work done by an external force to move a unit positive charge from R to P. $V_P - V_R = \frac{W_{RP}}{q} = \frac{U_P - U_R}{q}$

Like potential energy, it's the potential difference that is physically meaningful. If we again set the potential at infinity to be zero ($V_{\infty} = 0$), we can define the absolute potential at a point.

Electrostatic potential (V) at any point in an electric field is the work done by an external force in bringing a unit positive charge from infinity to that point without acceleration.

Potential due to a Point Charge

Let's find the potential at a point P, at a distance r from a single positive point charge Q located at the origin. To do this, we calculate the work done in bringing a unit positive test charge from infinity to P.

The electrostatic force on a unit positive charge at an intermediate point P' (at distance $r'$) is: $F = \frac{Q \times 1}{4 \pi \varepsilon_{0} r'^2}$ The work done to move it a small distance $dr'$ against this force is: $\Delta W = -\frac{Q}{4 \pi \varepsilon_{0} r'^2} \Delta r'$ The negative sign is there because to bring the charge closer, the displacement $\Delta r'$ is negative, but the work done against the repulsive force must be positive.

To find the total work done (which equals the potential V), we integrate this expression from infinity to r: $W = -\int_{\infty}^{r} \frac{Q}{4 \pi \varepsilon_{0} r'^2} dr' = \left[ \frac{Q}{4 \pi \varepsilon_{0} r'} \right]_{\infty}^{r} = \frac{Q}{4 \pi \varepsilon_{0} r} - 0$

So, the potential V at a distance r from a point charge Q is: $V(r) = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}$

This formula holds for any sign of Q:

• If Q is positive, V is positive.
• If Q is negative, V is negative.

As shown in Figure 2.4, the electric potential from a point charge decreases as $1/r$, while the electric field decreases more quickly, as $1/r^2$.

Given

• Source charge, $Q = 4 \times 10^{-7} \text{ C}$
• Distance, $r = 9 \text{ cm} = 0.09 \text{ m}$
• Test charge, $q = 2 \times 10^{-9} \text{ C}$
• Constant, $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \text{ Nm}^2 \text{C}^{-2}$

To Find

(a) Potential at point P, V (b) Work done, W

Formula

$V = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}$ $W = qV$

Solution

(a) Calculate the potential V

Substitute the given values into the formula for potential: $V = (9 \times 10^{9} \text{ Nm}^2 \text{C}^{-2}) \times \frac{4 \times 10^{-7} \text{ C}}{0.09 \text{ m}}$ $V = 4 \times 10^{4} \text{ V}$

Answer for part (a) = $4 \times 10^{4} \text{ V}$

(b) Calculate the work done W

Use the potential calculated in part (a) to find the work done: $W = qV = (2 \times 10^{-9} \text{ C}) \times (4 \times 10^{4} \text{ V})$ $W = 8 \times 10^{-5} \text{ J}$

The work done is path independent because the electrostatic force is a conservative force.

Answer for part (b) = $8 \times 10^{-5} \text{ J}$. No, the answer does not depend on the path.

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 at any point P is the sum of the potentials from each charge. $V = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q}{r_1} - \frac{q}{r_2} \right)$ where $r_1$ is the distance from +q to P and $r_2$ is the distance from -q to P.

For a point P that is far away from the dipole ($r \gg a$), we can use geometry and the binomial approximation to simplify the expression. The distances $r_1$ and $r_2$ can be approximated as: $\frac{1}{r_1} \approx \frac{1}{r} \left( 1 + \frac{a}{r} \cos\theta \right)$ $\frac{1}{r_2} \approx \frac{1}{r} \left( 1 - \frac{a}{r} \cos\theta \right)$ where $\theta$ is the angle between the position vector r of point P and the dipole moment vector p.

Substituting these into the potential equation, we get: $V = \frac{q}{4 \pi \varepsilon_{0}} \left[ \frac{1}{r} \left( 1 + \frac{a}{r} \cos\theta \right) - \frac{1}{r} \left( 1 - \frac{a}{r} \cos\theta \right) \right]$ $V = \frac{q}{4 \pi \varepsilon_{0}} \frac{2a \cos\theta}{r^2}$

Since the magnitude of the dipole moment is $p = q \times 2a$, the potential due to a dipole is: $V = \frac{p \cos\theta}{4 \pi \varepsilon_{0} r^2}$ This can also be written using the dot product, where $\hat{\mathbf{r}}$ is the unit vector along the position vector r: $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{\mathbf{p} \cdot \hat{\mathbf{r}}}{r^2} \quad (r \gg a)$

Key differences between potential from a single charge and a dipole:

1. Dependence on Angle: Dipole potential depends on both the distance r and the angle $\theta$.
2. Dependence on Distance: Dipole potential falls off as $1/r^2$, which is faster than the $1/r$ fall-off for a single point charge.
3. Special Cases:
   • On the dipole axis ($\theta = 0$ or $\pi$), $V = \pm \frac{p}{4 \pi \varepsilon_{0} r^2}$.
   • On the equatorial plane ($\theta = \pi/2$), the potential is zero.

Potential due to a System of Charges

The superposition principle applies to electrostatic potential. The total potential at a point P due to a system of charges ($q_1, q_2, \dots, q_n$) is the simple algebraic sum of the potentials due to each individual charge. $V = V_1 + V_2 + \dots + V_n$ $V = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q_1}{r_{1P}} + \frac{q_2}{r_{2P}} + \dots + \frac{q_n}{r_{nP}} \right)$ where $r_{1P}$ is the distance from charge $q_1$ to point P, and so on.

Potential due to a Uniformly Charged Spherical Shell

For a spherical shell with total charge q and radius R:

• Outside the shell ($r \ge R$): The potential is the same as if all the charge q were concentrated at the center. $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}$
• Inside the shell ($r < R$): The electric field inside the shell is zero. This means no work is done to move a charge inside. Therefore, the potential is constant everywhere inside the shell and is equal to the potential at the surface. $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R}$

Given

• First charge, $q_1 = 3 \times 10^{-8} \text{ C}$
• Second charge, $q_2 = -2 \times 10^{-8} \text{ C}$
• Separation distance = $15 \text{ cm} = 0.15 \text{ m}$
• Let the positive charge be at the origin (x=0) and the negative charge at x=15 cm.

To Find

The point(s) x where the total potential V is zero.

Formula

The total potential V at a point is the sum of the potentials from each charge: $V = V_1 + V_2 = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} \right)$ For V to be zero: $\frac{q_1}{r_1} + \frac{q_2}{r_2} = 0 \implies \frac{3 \times 10^{-8}}{r_1} + \frac{-2 \times 10^{-8}}{r_2} = 0$ $\frac{3}{r_1} = \frac{2}{r_2}$

Solution

There are two possible locations for the point P on the line.

Case 1: P is between the two charges. Let the distance from the positive charge ($q_1$) be x. Then the distance from the negative charge ($q_2$) is $(15-x)$. $r_1 = x \quad \text{and} \quad r_2 = 15-x$ Substituting into our condition: $\frac{3}{x} = \frac{2}{15-x}$ $3(15-x) = 2x$ $45 - 3x = 2x$ $5x = 45 \implies x = 9 \text{ cm}$

Case 2: P is on the extended line, to the right of the negative charge. Let the distance from the positive charge ($q_1$) be x. Then the distance from the negative charge ($q_2$) is $(x-15)$. $r_1 = x \quad \text{and} \quad r_2 = x-15$ Substituting into our condition: $\frac{3}{x} = \frac{2}{x-15}$ $3(x-15) = 2x$ $3x - 45 = 2x \implies x = 45 \text{ cm}$

(Note: There is no point to the left of the positive charge where the potential can be zero, as the potential from the closer, larger positive charge would always dominate the potential from the farther, smaller negative charge).

Final Answer The electric potential is zero at two points: 9 cm from the positive charge (between the charges) and 45 cm from the positive charge (on the side of the negative charge).

Equipotential Surfaces

An equipotential surface is a surface where the electric potential is the same at every point.

• For a single point charge q, the potential is $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}$. Since V is constant when r is constant, the equipotential surfaces are concentric spheres centered on the charge.
• For a uniform electric field, the equipotential surfaces are planes perpendicular to the electric field lines.

Key Property: The electric field E is always normal (perpendicular) to the equipotential surface at every point.

Relation between Field and Potential

The electric field and potential are intimately related. The electric field points in the direction where the potential decreases most rapidly. Its magnitude is given by the change in potential per unit displacement normal to the equipotential surface. $|\mathbf{E}| = -\frac{\delta V}{\delta l}$ where $\delta V$ is the potential difference between two closely spaced equipotential surfaces and $\delta l$ is the perpendicular distance between them. Since E points from higher to lower potential, $\delta V$ is negative, making $|\mathbf{E}|$ positive.

Potential Energy of a System of Charges

The potential energy of a system of charges is the total work done by an external force to assemble the charges from infinity to their final positions.

Two Charges

1. Bring charge $q_1$ from infinity to its position $\mathbf{r}_1$. No work is done because there is no external field. Work = 0.
2. Bring charge $q_2$ from infinity to its position $\mathbf{r}_2$. Now, we must do work against the electric field created by $q_1$. The potential at $\mathbf{r}_2$ due to $q_1$ is $V_1 = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_1}{r_{12}}$, where $r_{12}$ is the distance between them.
3. Work done on $q_2$ is $W_2 = q_2 V_1 = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_1 q_2}{r_{12}}$.

The total potential energy U of the two-charge system is the sum of the work done: $U = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_1 q_2}{r_{12}}$

• If the charges have the same sign ($q_1 q_2 > 0$), U is positive. Work must be done to bring them together against repulsion.
• If the charges have opposite signs ($q_1 q_2 < 0$), U is negative. The attractive force does work, so the external agent does negative work.

Three Charges

To find the potential energy of three charges ($q_1, q_2, q_3$), we sum the potential energy of each pair of charges: $U = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right)$

Potential Energy in an External Field

Here, we calculate the potential energy of a charge (or system of charges) placed in an electric field E (or potential V) that is created by external sources, not by the charges themselves.

Potential Energy of a Single Charge

The potential energy of a single charge q at a point r in an external potential $V(\mathbf{r})$ is the work done to bring that charge from infinity to the point r. $U = qV(\mathbf{r})$

Potential Energy of a System of Two Charges in an External Field

To find the total potential energy of two charges, $q_1$ and $q_2$, in an external field, we must consider three terms:

1. The work done to bring $q_1$ into the external field: $q_1 V(\mathbf{r}_1)$.
2. The work done to bring $q_2$ into the external field: $q_2 V(\mathbf{r}_2)$.
3. The work done to bring $q_2$ near $q_1$ (their mutual interaction): $\frac{1}{4 \pi \varepsilon_{0}} \frac{q_1 q_2}{r_{12}}$.

The total potential energy of the system is the sum of these: $U = q_1 V(\mathbf{r}_1) + q_2 V(\mathbf{r}_2) + \frac{1}{4 \pi \varepsilon_{0}} \frac{q_1 q_2}{r_{12}}$

Potential Energy of a Dipole in an External Field

Consider a dipole with moment p in a uniform external electric field E. The field exerts a torque $\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}$ on the dipole, which tends to align it with the field.

The work done by an external torque to rotate the dipole from an angle $\theta_0$ to $\theta_1$ is stored as potential energy. $W = \int_{\theta_0}^{\theta_1} pE \sin\theta d\theta = pE(\cos\theta_0 - \cos\theta_1)$ If we choose the reference position where potential energy is zero to be when the dipole is perpendicular to the field ($\theta_0 = \pi/2$), the potential energy at any angle $\theta$ becomes: $U(\theta) = pE(\cos(\pi/2) - \cos\theta) = -pE\cos\theta$ This can be written as a dot product: $U = -\mathbf{p} \cdot \mathbf{E}$

• Stable Equilibrium: Potential energy is minimum ($U = -pE$) when the dipole is aligned with the field ($\theta = 0$).
• Unstable Equilibrium: Potential energy is maximum ($U = +pE$) when the dipole is anti-aligned with the field ($\theta = \pi$).

Electrostatics of Conductors

Conductors contain mobile charge carriers (electrons in metals). In a static situation (no current flowing), they have several important properties:

1. Inside a conductor, the electrostatic field is zero. If there were a field, the free charges would move until they rearranged themselves to cancel it out.
2. At the surface of a charged conductor, the electrostatic field must be normal to the surface. If there were a tangential component, charges on the surface would move, which doesn't happen in a static situation.
3. The interior of a conductor has no excess charge. Any net charge given to a conductor resides entirely on its outer surface. This is a consequence of Gauss's law.
4. Electrostatic potential is constant throughout the volume of the conductor and on its surface. Since E=0 inside, no work is done moving a charge from one point to another within the conductor.
5. The electric field at the surface of a charged conductor is given by $\mathbf{E} = \frac{\sigma}{\varepsilon_0} \hat{\mathbf{n}}$, where $\sigma$ is the surface charge density and $\hat{\mathbf{n}}$ is a unit vector normal to the surface pointing outwards.
6. Electrostatic Shielding: The electric field inside any cavity within a conductor is zero, regardless of the charges on the conductor or external fields. This is used to protect sensitive electronic equipment from external electrical interference.

Dielectrics and Polarisation

Dielectrics are insulating materials that do not have free charges like conductors. However, when placed in an external electric field, they can be polarised.

Molecules can be classified as non-polar or polar:

• Non-polar molecules (like $O_2$, $H_2$): The centers of positive and negative charge coincide, so they have no permanent dipole moment. In an external E-field, the charges are slightly displaced, creating an induced dipole moment.
• Polar molecules (like $H_2O$, HCl): The centers of positive and negative charge are naturally separated, giving them a permanent dipole moment. In the absence of an E-field, these dipoles are randomly oriented. When an external E-field is applied, they tend to align with the field.

In both cases, the dielectric as a whole develops a net dipole moment in the presence of an external field. This phenomenon is called polarisation.

When a dielectric slab is placed in an external field $\mathbf{E}_0$, the polarisation creates an internal electric field that opposes the external field. This results in a net electric field inside the dielectric that is weaker than the original external field.

Capacitors and Capacitance

A capacitor is a device for storing electrical energy, consisting of two conductors separated by an insulator (a dielectric).

When the conductors are given charges Q and -Q, a potential difference V develops between them. The charge Q is found to be directly proportional to the potential difference V. The constant of proportionality is called the capacitance (C). $C = \frac{Q}{V}$

• Capacitance is a measure of a capacitor's ability to store charge. It depends only on the geometry (shape, size, separation) of the conductors and the dielectric material between them.
• The SI unit of capacitance is the farad (F). $1 \text{ Farad} = 1 \text{ Coulomb/Volt}$. The farad is a very large unit; more common units are the microfarad ($\mu\text{F} = 10^{-6} \text{ F}$) and picofarad ($\text{pF} = 10^{-12} \text{ F}$).

A capacitor with a large capacitance can store a large amount of charge at a relatively low potential difference.

The Parallel Plate Capacitor

This is the simplest type of capacitor, made of two large, parallel conducting plates, each of area A, separated by a small distance d.

If the plates have charges Q and -Q, the surface charge density is $\sigma = Q/A$. The electric field E between the plates is uniform and is given by: $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$ The potential difference V between the plates is: $V = Ed = \frac{Qd}{\varepsilon_0 A}$ The capacitance C is therefore: $C = \frac{Q}{V} = \frac{Q}{Qd / \varepsilon_0 A}$ $C = \frac{\varepsilon_0 A}{d}$

This shows that capacitance is increased by having larger plate area (A) and a smaller separation (d).

Effect of Dielectric on Capacitance

If a dielectric material is inserted to completely fill the space between the capacitor plates, it becomes polarized. This reduces the electric field between the plates from $E_0$ (in vacuum) to $E = E_0/K$, where K is the dielectric constant of the material.

The potential difference also reduces: $V = V_0/K$. Since capacitance is $C = Q/V$, the new capacitance is: $C = \frac{Q}{V_0/K} = K \frac{Q}{V_0} = K C_0$ where $C_0$ is the capacitance with a vacuum between the plates.

The dielectric constant K is a dimensionless factor ($K > 1$) that tells you how much the capacitance is increased by the presence of the dielectric. $K = \frac{C}{C_0}$

Combination of Capacitors

Capacitors can be combined in circuits in two basic ways: in series or in parallel.

Capacitors in Series

When capacitors are connected end-to-end, they are in series.

• Charge: The charge Q is the same on each capacitor.
• Voltage: The total potential difference V across the combination is the sum of the individual potential differences ($V = V_1 + V_2 + \dots$).

The equivalent capacitance $C_{eq}$ is given by: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots$

Capacitors in Parallel

When capacitors are connected across the same two points, they are in parallel.

• Voltage: The potential difference V is the same across each capacitor.
• Charge: The total charge Q stored is the sum of the charges on each capacitor ($Q = Q_1 + Q_2 + \dots$).

The equivalent capacitance $C_{eq}$ is given by: $C_{eq} = C_1 + C_2 + C_3 + \dots$

Given

• Capacitances: $C_1 = C_2 = C_3 = C_4 = 10 \mu\text{F}$
• Supply Voltage, $V = 500 \text{ V}$
• $C_1, C_2, C_3$ are in series. This combination is in parallel with $C_4$.

To Find

(a) The equivalent capacitance of the network, C (b) The charge on each capacitor ($Q_1, Q_2, Q_3, Q_4$)

Formula

• For series combination: $\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$
• For parallel combination: $C = C' + C_4$
• Charge on a capacitor: $Q = CV$

Solution

(a) Calculate the equivalent capacitance

First, find the equivalent capacitance $C'$ for the three capacitors in series ($C_1, C_2, C_3$): $\frac{1}{C'} = \frac{1}{10 \mu\text{F}} + \frac{1}{10 \mu\text{F}} + \frac{1}{10 \mu\text{F}} = \frac{3}{10 \mu\text{F}}$ $C' = \frac{10}{3} \mu\text{F}$ Next, this combination ($C'$) is in parallel with $C_4$. The total equivalent capacitance C is: $C = C' + C_4 = \frac{10}{3} \mu\text{F} + 10 \mu\text{F} = \frac{10+30}{3} \mu\text{F} = \frac{40}{3} \mu\text{F}$ $C \approx 13.3 \mu\text{F}$

Answer for part (a) = $13.3 \mu\text{F}$

(b) Calculate the charge on each capacitor

The potential difference across the parallel branch containing $C_4$ is 500 V. So, the charge on $C_4$ is: $Q_4 = C_4 V = (10 \times 10^{-6} \text{ F}) \times (500 \text{ V}) = 5 \times 10^{-3} \text{ C}$ The potential difference across the series branch ($C_1, C_2, C_3$) is also 500 V. The charge on the equivalent capacitor $C'$ is: $Q' = C' V = \left(\frac{10}{3} \times 10^{-6} \text{ F}\right) \times (500 \text{ V}) = \frac{5000}{3} \times 10^{-6} \text{ C} \approx 1.7 \times 10^{-3} \text{ C}$ Since $C_1, C_2,$ and $C_3$ are in series, the charge on each of them is the same, and it is equal to the charge $Q'$ on their equivalent combination. $Q_1 = Q_2 = Q_3 = Q' \approx 1.7 \times 10^{-3} \text{ C}$

Answer for part (b) = Charge on $C_1, C_2, C_3$ is $1.7 \times 10^{-3} \text{ C}$. Charge on $C_4$ is $5.0 \times 10^{-3} \text{ C}$.

Energy Stored in a Capacitor

The work done to charge a capacitor is stored as electrostatic potential energy in the electric field between its plates.

To charge a capacitor from 0 to a final charge Q, work must be done to move charge from one plate to the other against the building potential difference. The total work done, W, which is equal to the stored energy U, is: $U = W = \frac{Q^2}{2C}$ Using the relation $Q=CV$, we can write this in two other equivalent forms: $U = \frac{1}{2} C V^2 = \frac{1}{2} Q V$

Energy Density of the Electric Field

The energy stored in a capacitor can be thought of as being stored in the electric field itself. For a parallel plate capacitor, the energy is: $U = \frac{1}{2} C V^2 = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right) (Ed)^2 = \frac{1}{2} \varepsilon_0 E^2 (Ad)$ The term Ad is the volume of the space between the plates. The energy density (u), or energy stored per unit volume, is: $u = \frac{U}{\text{Volume}} = \frac{1}{2} \varepsilon_0 E^2$ This formula is general and applies to any electric field, not just the one in a parallel plate capacitor.

Given

• Capacitance, $C = 900 \text{ pF} = 900 \times 10^{-12} \text{ F}$
• Initial voltage, $V = 100 \text{ V}$
• Second capacitor, $C_2 = 900 \text{ pF}$

To Find

(a) Initial energy stored, $U_{initial}$ (b) Final energy stored in the system, $U_{final}$

Formula

• Energy stored in a capacitor: $U = \frac{1}{2}CV^2$
• Charge on a capacitor: $Q = CV$

Solution

(a) Calculate the initial energy stored

$U_{initial} = \frac{1}{2}CV^2 = \frac{1}{2} (900 \times 10^{-12} \text{ F}) (100 \text{ V})^2$ $U_{initial} = \frac{1}{2} (900 \times 10^{-12}) (10000) = 4.5 \times 10^{-6} \text{ J}$

Answer for part (a) = $4.5 \times 10^{-6} \text{ J}$

(b) Calculate the final energy stored

First, find the initial charge on the first capacitor: $Q = CV = (900 \times 10^{-12} \text{ F}) (100 \text{ V}) = 9 \times 10^{-8} \text{ C}$ When this charged capacitor is connected to an identical uncharged capacitor, the total charge Q is conserved and distributes equally between the two. The charge on each capacitor becomes $Q' = Q/2$. The new potential difference across each capacitor, $V'$, will be half the original voltage, since $V' = Q'/C = (Q/2)/C = V/2 = 50 \text{ V}$.

The final total energy of the system is the sum of the energies in the two capacitors: $U_{final} = U_1' + U_2' = \frac{1}{2}C(V')^2 + \frac{1}{2}C(V')^2 = C(V')^2$ $U_{final} = (900 \times 10^{-12} \text{ F}) (50 \text{ V})^2 = (900 \times 10^{-12}) (2500) = 2.25 \times 10^{-6} \text{ J}$

Alternatively, using charge: $U_{final} = 2 \times \left( \frac{1}{2} \frac{(Q')^2}{C} \right) = \frac{(Q/2)^2}{C} = \frac{Q^2}{4C} = \frac{1}{2} U_{initial}$ $U_{final} = \frac{1}{2} (4.5 \times 10^{-6} \text{ J}) = 2.25 \times 10^{-6} \text{ J}$

Answer for part (b) = $2.25 \times 10^{-6} \text{ J}$