Chapter Notes

Introduction to Electrostatics

Have you ever felt a small spark when taking off a sweater in dry weather, or seen lightning flash across the sky during a storm? These are common examples of static electricity. The study of these phenomena, which involve stationary electric charges, is called electrostatics. It explores the forces, fields, and potentials that arise from charges that are not in motion. The reason we experience these sparks or shocks is due to the discharge of electric charges that have built up on surfaces, often through rubbing.

Electric Charge

The concept of electric charge dates back to ancient Greece, around 600 BC, when Thales of Miletus observed that amber rubbed with wool could attract light objects. The word "electricity" itself comes from elektron, the Greek word for amber.

Through many experiments, scientists discovered that there are two types of electric charge. The fundamental rule governing their interaction is:

• Like charges repel each other.
• Unlike charges attract each other.

For example, two glass rods rubbed with silk will repel each other, but a glass rod will attract the silk cloth it was rubbed with. This property that distinguishes the two types of charges is called the polarity of charge.

Benjamin Franklin named these two types of charges positive and negative.

• By convention, the charge on a glass rod rubbed with silk is positive.
• The charge on a plastic rod rubbed with fur is negative.

An object with an electric charge is said to be electrified or charged. An object with no net charge is electrically neutral.

How do objects get charged?

All matter is made of atoms, which contain positive charges (protons) and negative charges (electrons). Usually, these charges are balanced, so the object is neutral. Charging an object involves transferring electrons from one body to another.

• A body becomes positively charged when it loses electrons.
• A body becomes negotively charged when it gains electrons.

When a glass rod is rubbed with silk, electrons move from the rod to the silk. The rod becomes positively charged, and the silk becomes negatively charged. No new charge is created; it is simply redistributed.

A gold-leaf electroscope is a simple device used to detect the presence of charge on an object. When a charged object touches the metal knob, the charge spreads to the thin gold leaves at the bottom, causing them to repel each other and diverge.

Conductors and Insulators

Materials can be classified based on how easily they allow electric charge to move through them.

• Conductors are materials that allow electricity to pass through them easily. They have mobile electric charges (usually electrons) that are free to move. Examples include metals, the human body, and the Earth. When charge is given to a conductor, it quickly spreads over its entire surface.
• Insulators are materials that offer high resistance to the passage of electricity. Their charges are not free to move. Examples include glass, plastic, wood, and nylon. If charge is put on an insulator, it stays in the same place.

Basic Properties of Electric Charge

Electric charge has three fundamental properties: additivity, conservation, and quantisation. For calculations involving forces between charges, we often treat charged bodies as point charges if their sizes are much smaller than the distance between them.

Additivity of Charges

Electric charge is a scalar quantity, meaning it has magnitude but no direction. The total charge of a system is simply the algebraic sum of all the individual charges in it. You must use the proper signs (positive or negative) when adding them.

If a system contains charges $q_1, q_2, q_3, \ldots, q_n$, the total charge $Q$ is: $Q = q_1 + q_2 + q_3 + \ldots + q_n$

Charge is Conserved

The law of conservation of charge states that the total charge of an isolated system always remains constant. When you rub two objects together, charge is transferred from one to the other, but the net charge of the system (the two objects) does not change. Charge cannot be created or destroyed, only moved around.

Even in nuclear processes where particles are created, charge is conserved. For instance, when a neutron (no charge) decays, it turns into a proton (positive charge) and an electron (negative charge). The net charge is zero both before and after the process.

Quantisation of Charge

Experimentally, it's found that all free charges are integral multiples of a basic unit of charge, denoted by e. This property is called the quantisation of charge.

The charge q on any body is given by the formula: $q = ne$

Where n is an integer ($0, \pm1, \pm2, \ldots$) and e is the elementary charge, which is the magnitude of the charge on a single electron or proton.

• Charge on an electron = $-e$
• Charge on a proton = $+e$

The value of the elementary charge in SI units is: $e = 1.602192 \times 10^{-19} \text{ C}$ The SI unit of charge is the coulomb (C).

Given

• Number of electrons transferred per second, $n_{sec} = 10^9$
• Total charge to be accumulated, $Q = 1$ C
• Elementary charge, $e = 1.6 \times 10^{-19}$ C

To Find

The time required, $t$, to accumulate a charge of 1 C.

Formula

1. Charge transferred per second, $q_{sec} = n_{sec} \times e$
2. Total time, $t = \frac{Q}{q_{sec}}$

Solution

First, calculate the charge transferred in one second: $q_{sec} = 10^9 \times (1.6 \times 10^{-19} \text{ C}) = 1.6 \times 10^{-10} \text{ C/s}$ Now, calculate the total time required to accumulate 1 C: $t = \frac{1 \text{ C}}{1.6 \times 10^{-10} \text{ C/s}} = 6.25 \times 10^9 \text{ s}$ To convert this to years: $t = \frac{6.25 \times 10^9 \text{ s}}{365 \times 24 \times 3600 \text{ s/year}} \approx 198 \text{ years}$

Final Answer It would take approximately 198 years to accumulate a charge of 1 C. This shows that one coulomb is a very large unit of charge for practical purposes.

Given

• Mass of water in a cup (assumed), $m = 250$ g
• Molecular mass of water ($H_2O$), $M = 18$ g/mol
• Avogadro's number, $N_A = 6.02 \times 10^{23}$ molecules/mol
• A water molecule ($H_2O$) has 2 Hydrogen atoms (1 proton, 1 electron each) and 1 Oxygen atom (8 protons, 8 electrons). Total = 10 protons and 10 electrons.

To Find

The total positive and negative charge in the cup of water.

Formula

1. Number of molecules, $N = (\frac{m}{M}) \times N_A$
2. Total charge, $Q = N \times (\text{number of protons or electrons per molecule}) \times e$

Solution

First, find the number of molecules in 250 g of water: $N = \left(\frac{250 \text{ g}}{18 \text{ g/mol}}\right) \times (6.02 \times 10^{23} \text{ molecules/mol})$ The total positive charge (from 10 protons per molecule) and total negative charge (from 10 electrons per molecule) have the same magnitude. Let's calculate this magnitude: $Q = \left(\frac{250}{18} \times 6.02 \times 10^{23}\right) \times 10 \times (1.6 \times 10^{-19} \text{ C})$ $Q \approx 1.34 \times 10^7 \text{ C}$

Final Answer A cup of water contains approximately $1.34 \times 10^7$ C of positive charge and $1.34 \times 10^7$ C of negative charge.

Coulomb's Law

Coulomb's Law is a fundamental law that quantifies the electrostatic force between two stationary point charges. It states that the force is:

1. Directly proportional to the product of the magnitudes of the two charges.
2. Inversely proportional to the square of the distance between them.
3. Acts along the line connecting the two charges.

The magnitude of the force F between two point charges $q_1$ and $q_2$ separated by a distance r in a vacuum is given by: $F = k \frac{|q_1 q_2|}{r^2}$ Here, k is the electrostatic force constant. For convenience, k is usually written as $k = \frac{1}{4\pi\varepsilon_0}$, where $\varepsilon_0$ is the permittivity of free space.

So, Coulomb's Law is written as: $F = \frac{1}{4\pi\varepsilon_0} \frac{|q_1 q_2|}{r^2}$

In SI units:

• $\varepsilon_0 \approx 8.854 \times 10^{-12} \text{ C}^2 \text{N}^{-1} \text{m}^{-2}$
• $k \approx 9 \times 10^9 \text{ N m}^2 \text{C}^{-2}$

Coulomb's Law in Vector Form

Force is a vector, so it's more complete to write Coulomb's law in vector form. Let $\mathbf{r}_1$ and $\mathbf{r}_2$ be the position vectors of charges $q_1$ and $q_2$. The force on charge $q_2$ due to $q_1$, denoted by $\mathbf{F}_{21}$, is: $\mathbf{F}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{21}^2} \hat{\mathbf{r}}_{21}$ where $\mathbf{r}_{21} = \mathbf{r}_2 - \mathbf{r}_1$ is the vector from $q_1$ to $q_2$, and $\hat{\mathbf{r}}_{21}$ is the unit vector in that direction.

• If $q_1$ and $q_2$ have the same sign, the product $q_1q_2$ is positive, and $\mathbf{F}_{21}$ is in the direction of $\hat{\mathbf{r}}_{21}$ (repulsion).
• If $q_1$ and $q_2$ have opposite signs, the product $q_1q_2$ is negative, and $\mathbf{F}_{21}$ is in the direction opposite to $\hat{\mathbf{r}}_{21}$ (attraction).

The force on $q_1$ due to $q_2$ is $\mathbf{F}_{12} = -\mathbf{F}_{21}$, which is consistent with Newton's third law.

Given

• Distance between electron and proton, $r = 1 \text{ Å} = 10^{-10}$ m
• Mass of proton, $m_p = 1.67 \times 10^{-27}$ kg
• Mass of electron, $m_e = 9.11 \times 10^{-31}$ kg
• Charge, $e = 1.6 \times 10^{-19}$ C
• $k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{C}^{-2}$
• Gravitational constant, $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}$

To Find

(a) Ratio of magnitudes of electric force ($F_e$) to gravitational force ($F_G$) for (i) an electron and a proton, and (ii) two protons. (b) Acceleration of the electron ($a_e$) and proton ($a_p$) when they are $1 \text{ Å}$ apart.

Formula

$F_e = k \frac{e^2}{r^2}$ $F_G = G \frac{m_1 m_2}{r^2}$ $F = ma \implies a = F/m$

Solution

(a) (i) Ratio for an electron and a proton The ratio of the magnitudes of the forces is independent of the distance r: $\left|\frac{F_e}{F_G}\right| = \frac{k e^2 / r^2}{G m_p m_e / r^2} = \frac{k e^2}{G m_p m_e}$ $\left|\frac{F_e}{F_G}\right| = \frac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{(6.67 \times 10^{-11})(1.67 \times 10^{-27})(9.11 \times 10^{-31})} \approx 2.4 \times 10^{39}$

(a) (ii) Ratio for two protons $\left|\frac{F_e}{F_G}\right| = \frac{k e^2}{G m_p m_p}$ $\left|\frac{F_e}{F_G}\right| = \frac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{(6.67 \times 10^{-11})(1.67 \times 10^{-27})^2} \approx 1.3 \times 10^{36}$ This shows that electrical forces are enormously stronger than gravitational forces.

(b) Accelerations of electron and proton First, calculate the magnitude of the electric force between them at $r = 10^{-10}$ m: $|\mathbf{F}| = k \frac{e^2}{r^2} = (9 \times 10^9 \text{ Nm}^2/\text{C}^2) \times \frac{(1.6 \times 10^{-19} \text{ C})^2}{(10^{-10} \text{ m})^2}$ $|\mathbf{F}| = 2.3 \times 10^{-8} \text{ N}$ Now, find the accelerations using $a = F/m$: Acceleration of the electron: $a_e = \frac{2.3 \times 10^{-8} \text{ N}}{9.11 \times 10^{-31} \text{ kg}} = 2.5 \times 10^{22} \text{ m/s}^2$ Acceleration of the proton: $a_p = \frac{2.3 \times 10^{-8} \text{ N}}{1.67 \times 10^{-27} \text{ kg}} = 1.4 \times 10^{19} \text{ m/s}^2$

Final Answer The ratios show the immense strength of the electric force. The accelerations are huge, indicating that for atomic particles, gravity's effect is negligible compared to the electrostatic force.

Forces between Multiple Charges

What if there are more than two charges? To find the force on one charge due to several others, we use the principle of superposition.

This principle states that the total force on any given charge is the vector sum of the individual forces exerted on it by all other charges. The force between any two charges is calculated using Coulomb's law and is unaffected by the presence of other charges.

For a system of charges $q_1, q_2, \ldots, q_n$, the total force $\mathbf{F}_1$ on charge $q_1$ is: $\mathbf{F}_1 = \mathbf{F}_{12} + \mathbf{F}_{13} + \ldots + \mathbf{F}_{1n}$ $\mathbf{F}_1 = \frac{1}{4\pi\varepsilon_0} \sum_{i=2}^{n} \frac{q_1 q_i}{r_{1i}^2} \hat{\mathbf{r}}_{1i}$

Given

• Three charges, $q_1 = q_2 = q_3 = q$, at the vertices of an equilateral triangle.
• Side length of the triangle = $l$.
• A charge $Q$ is placed at the centroid O.

To Find

The net force on charge $Q$.

Solution

The centroid of an equilateral triangle is equidistant from all three vertices. Let this distance be r. The distance from a vertex to the centroid (AO, BO, CO) is $r = l/\sqrt{3}$.

The force on $Q$ due to the charge $q$ at vertex A is $\mathbf{F}_1$. Its magnitude is: $F_1 = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{(l/\sqrt{3})^2} = \frac{3}{4\pi\varepsilon_0} \frac{Qq}{l^2}$ This force is directed along AO (repulsive).

Similarly, the forces on $Q$ due to charges at B and C, $\mathbf{F}_2$ and $\mathbf{F}_3$, have the same magnitude and are directed along BO and CO, respectively.

The three force vectors $\mathbf{F}_1, \mathbf{F}_2, \mathbf{F}_3$ are equal in magnitude and are $120^\circ$ apart. By symmetry, the vector sum of these three forces must be zero. If you place three equal vectors at $120^\circ$ to each other, their resultant is zero.

$\mathbf{F}_{net} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = 0$

Final Answer The force on the charge $Q$ placed at the centroid is zero.

Electric Field

Instead of thinking about forces acting at a distance, it's useful to introduce the concept of a field. A source charge Q creates an electric field in the space around it. When another charge q (a test charge) is placed in this field, it experiences a force.

The electric field $\mathbf{E}$ at a point is defined as the electrostatic force $\mathbf{F}$ experienced by a small positive test charge q placed at that point, divided by the magnitude of the test charge. $\mathbf{E} = \lim_{q \to 0} \frac{\mathbf{F}}{q}$ The test charge q must be negligibly small so it doesn't disturb the source charge(s).

The SI unit for electric field is newtons per coulomb (N/C).

For a single source charge Q, the electric field at a distance r is: $\mathbf{E} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat{\mathbf{r}}$

• If Q is positive, $\mathbf{E}$ points radially outward.
• If Q is negative, $\mathbf{E}$ points radially inward.

The force on any charge q placed in an electric field $\mathbf{E}$ is given by: $\mathbf{F} = q\mathbf{E}$

Electric Field due to a System of Charges

Just like with forces, the electric field from multiple source charges follows the superposition principle. The total electric field at a point is the vector sum of the electric fields from each individual charge. $\mathbf{E}(\mathbf{r}) = \mathbf{E}_1(\mathbf{r}) + \mathbf{E}_2(\mathbf{r}) + \ldots + \mathbf{E}_n(\mathbf{r})$ $\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{n} \frac{q_i}{r_{iP}^2} \hat{\mathbf{r}}_{iP}$

Given

• $q_1 = +10^{-8}$ C
• $q_2 = -10^{-8}$ C
• Distance between charges = 0.1 m
• Point A is the midpoint between the charges.
• Point B is 0.05 m from $q_1$ and 0.15 m from $q_2$.
• Point C is 0.1 m from both $q_1$ and $q_2$, forming an equilateral triangle.

To Find

The electric field at points A, B, and C.

Formula

$\mathbf{E} = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \hat{\mathbf{r}}$

Solution

At Point A (Midpoint): The distance from A to both $q_1$ and $q_2$ is $r = 0.05$ m.

• Field due to $q_1$ ($\mathbf{E}_{1A}$): Points to the right (away from positive charge). $E_{1A} = (9 \times 10^9) \frac{10^{-8}}{(0.05)^2} = 3.6 \times 10^4 \text{ N/C}$
• Field due to $q_2$ ($\mathbf{E}_{2A}$): Also points to the right (towards negative charge). $E_{2A} = (9 \times 10^9) \frac{|-10^{-8}|}{(0.05)^2} = 3.6 \times 10^4 \text{ N/C}$ The total field at A is the sum of these two vectors, which are in the same direction. $E_A = E_{1A} + E_{2A} = 7.2 \times 10^4 \text{ N/C}$ The direction is to the right.

At Point B:

• Field due to $q_1$ ($\mathbf{E}_{1B}$): Points to the left (away from $q_1$). Distance is 0.05 m. $E_{1B} = (9 \times 10^9) \frac{10^{-8}}{(0.05)^2} = 3.6 \times 10^4 \text{ N/C}$
• Field due to $q_2$ ($\mathbf{E}_{2B}$): Points to the right (towards $q_2$). Distance is 0.15 m. $E_{2B} = (9 \times 10^9) \frac{|-10^{-8}|}{(0.15)^2} = 4 \times 10^3 \text{ N/C}$ The total field at B is the vector difference. $E_B = E_{1B} - E_{2B} = 3.6 \times 10^4 - 0.4 \times 10^4 = 3.2 \times 10^4 \text{ N/C}$ The direction is to the left (the direction of the larger vector).

At Point C: The distance from C to both charges is $r = 0.10$ m. The magnitudes of the fields are equal. $E_{1C} = E_{2C} = (9 \times 10^9) \frac{10^{-8}}{(0.10)^2} = 9 \times 10^3 \text{ N/C}$

• $\mathbf{E}_{1C}$ points away from $q_1$.
• $\mathbf{E}_{2C}$ points towards $q_2$. The angle between these vectors is $120^\circ$. When we add them, the vertical components cancel out. The horizontal components (pointing to the right) add up. The angle inside the parallelogram is $60^\circ$ ($\pi/3$ radians). $E_C = E_{1C} \cos(60^\circ) + E_{2C} \cos(60^\circ)$ $E_C = (9 \times 10^3)(\frac{1}{2}) + (9 \times 10^3)(\frac{1}{2}) = 9 \times 10^3 \text{ N/C}$ The direction is to the right.

Electric Field Lines

Electric field lines are a way to visualize electric fields. They are imaginary curves drawn in a region of space so that the tangent to the curve at any point gives the direction of the electric field at that point. They are also known as lines of force.

Properties of Electric Field Lines:

• Direction: The tangent to a field line at any point gives the direction of the electric field $\mathbf{E}$ at that point. Arrows on the lines indicate this direction.
• Strength: The density of the field lines (how close they are to each other) represents the magnitude of the electric field. Field lines are close together where the field is strong and far apart where it is weak.
• Origin and Termination: Field lines originate from positive charges and terminate on negative charges. For a single charge, they may start or end at infinity.
• Continuity: Field lines are continuous curves and do not have sudden breaks in a charge-free region.
• No Crossing: Two field lines can never cross each other. If they did, it would mean the electric field has two different directions at the same point, which is impossible.
• No Closed Loops: Electrostatic field lines do not form closed loops. This is because the electrostatic force is a conservative force.

Electric Flux

Electric flux is a measure of the "flow" of the electric field through a given area. Imagine holding a ring in a flowing stream; the amount of water passing through the ring depends on its area and how it's oriented relative to the flow. Electric flux is an analogous concept.

For a small planar area element $\Delta\mathbf{S}$ in a uniform electric field $\mathbf{E}$, the electric flux $\Delta\phi$ is defined as the dot product of the electric field and the area vector: $\Delta\phi = \mathbf{E} \cdot \Delta\mathbf{S} = E \Delta S \cos\theta$ where $\theta$ is the angle between the electric field vector $\mathbf{E}$ and the normal to the surface (the direction of $\Delta\mathbf{S}$).

• The area element $\Delta\mathbf{S}$ is a vector whose magnitude is the area $\Delta S$ and whose direction is perpendicular (normal) to the surface.
• For a closed surface, the direction of the area vector is always taken to be the outward normal.
• The SI unit of electric flux is $\text{N m}^2 \text{C}^{-1}$.

The total flux $\phi$ through an entire surface S is the sum (or integral) of the flux through all the small area elements that make up the surface: $\phi \approx \sum \mathbf{E} \cdot \Delta\mathbf{S}$

Electric Dipole

An electric dipole is a pair of equal and opposite point charges, +q and -q, separated by a small distance, 2a.

The electric dipole moment ($\mathbf{p}$) is a vector quantity that characterizes the dipole.

• Magnitude: $p = q \times (2a)$
• Direction: From the negative charge (-q) to the positive charge (+q).

The SI unit for dipole moment is coulomb-meter (C m).

The Field of an Electric Dipole

The electric field of a dipole decreases more rapidly with distance than that of a single point charge. At large distances ($r \gg a$), the field strength is proportional to $1/r^3$.

(i) For points on the axis (Axial Line) The electric field at a point P on the axis of the dipole, at a distance r from its center, is: $\mathbf{E} = \frac{1}{4\pi\varepsilon_0} \frac{2\mathbf{p}r}{(r^2 - a^2)^2}$ For large distances ($r \gg a$): $\mathbf{E} \approx \frac{1}{4\pi\varepsilon_0} \frac{2\mathbf{p}}{r^3}$ The field is in the same direction as the dipole moment $\mathbf{p}$.

(ii) For points on the equatorial plane The electric field at a point P on the plane perpendicular to the dipole axis and passing through its center, at a distance r from the center, is: $\mathbf{E} = -\frac{1}{4\pi\varepsilon_0} \frac{\mathbf{p}}{(r^2 + a^2)^{3/2}}$ For large distances ($r \gg a$): $\mathbf{E} \approx -\frac{1}{4\pi\varepsilon_0} \frac{\mathbf{p}}{r^3}$ The field is in the direction opposite to the dipole moment $\mathbf{p}$.

Given

• Charge, $q = 10 \mu\text{C} = 10^{-5}$ C
• Separation, $2a = 5.0 \text{ mm} = 5 \times 10^{-3}$ m (so $a = 2.5 \times 10^{-3}$ m)
• Distance to points P and Q, $r = 15 \text{ cm} = 0.15$ m

To Find

(a) Electric field at axial point P, $\mathbf{E}_P$ (b) Electric field at equatorial point Q, $\mathbf{E}_Q$

Formula

Since $r=15$ cm is much larger than $a=0.25$ cm ($r/a = 60$), we can use the simplified formulas for large distances. Dipole moment magnitude: $p = q \times (2a)$ Axial field: $E_{axial} = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3}$ Equatorial field: $E_{equatorial} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3}$

Solution

First, calculate the magnitude of the dipole moment: $p = (10^{-5} \text{ C}) \times (5 \times 10^{-3} \text{ m}) = 5 \times 10^{-8} \text{ C m}$ The direction of $\mathbf{p}$ is from the negative to the positive charge.

(a) Field at axial point P The point P is on the side of the positive charge, so the field will be in the same direction as the dipole moment. $E_P = (9 \times 10^9 \text{ N m}^2 \text{C}^{-2}) \times \frac{2 \times (5 \times 10^{-8} \text{ C m})}{(0.15 \text{ m})^3}$ $E_P = (9 \times 10^9) \times \frac{10 \times 10^{-8}}{0.003375} = 2.66 \times 10^5 \text{ N/C}$

Answer for part (a) = $2.7 \times 10^5 \text{ N/C}$ (using the more precise calculation from the text), directed along the dipole axis from negative to positive charge.

(b) Field at equatorial point Q The field at an equatorial point is directed opposite to the dipole moment. $E_Q = (9 \times 10^9 \text{ N m}^2 \text{C}^{-2}) \times \frac{5 \times 10^{-8} \text{ C m}}{(0.15 \text{ m})^3}$ $E_Q = (9 \times 10^9) \times \frac{5 \times 10^{-8}}{0.003375} = 1.33 \times 10^5 \text{ N/C}$

Answer for part (b) = $1.33 \times 10^5 \text{ N/C}$, directed opposite to the dipole moment (parallel to the axis).

Dipole in a Uniform External Field

Consider a dipole with moment $\mathbf{p}$ placed in a uniform external electric field $\mathbf{E}$.

• The positive charge +q experiences a force $q\mathbf{E}$.
• The negative charge -q experiences a force $-q\mathbf{E}$.

Since the field is uniform, the net force on the dipole is zero ($q\mathbf{E} - q\mathbf{E} = 0$). The dipole will not accelerate linearly.

However, because the forces act at different points, they create a torque that causes the dipole to rotate. The magnitude of this torque ($\tau$) is: $\tau = pE\sin\theta$ where $\theta$ is the angle between the dipole moment $\mathbf{p}$ and the electric field $\mathbf{E}$.

In vector form, the torque is given by the cross product: $\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}$ This torque tends to align the dipole with the electric field. The torque is maximum when $\theta = 90^\circ$ and zero when the dipole is aligned with the field ($\theta = 0^\circ$).

If the electric field is non-uniform, the net force on the dipole is generally non-zero, and it will experience both a torque and a net force.

Continuous Charge Distribution

For many practical situations, like the charge on a metal sheet, it's more convenient to work with a continuous charge distribution rather than individual point charges. We define charge densities for this purpose.

• Linear Charge Density ($\lambda$): Charge per unit length. Used for objects like a thin wire. $\lambda = \frac{\Delta Q}{\Delta l}$ Unit: C/m

• Surface Charge Density ($\sigma$): Charge per unit area. Used for objects like a thin sheet. $\sigma = \frac{\Delta Q}{\Delta S}$ Unit: C/m$^2$

• Volume Charge Density ($\rho$): Charge per unit volume. Used for three-dimensional objects. $\rho = \frac{\Delta Q}{\Delta V}$ Unit: C/m$^3$

To find the electric field from a continuous charge distribution, we can imagine it is made of infinitely many small charge elements ($\rho\Delta V$). We calculate the field from each element using Coulomb's law and then sum (integrate) all their contributions using the superposition principle.

Gauss's Law

Gauss's Law provides a powerful relationship between the electric flux through a closed surface and the net charge enclosed within that surface.

The law states: The total electric flux ($\phi$) through any closed surface (called a Gaussian surface) is equal to the net charge enclosed ($q_{enc}$) divided by the permittivity of free space ($\varepsilon_0$).

$\phi = \oint \mathbf{E} \cdot d\mathbf{S} = \frac{q_{enc}}{\varepsilon_0}$

Key Points about Gauss's Law:

• It is true for any closed surface, regardless of its shape or size.
• The charge $q_{enc}$ is the algebraic sum of all charges inside the surface.
• The electric field $\mathbf{E}$ on the left side of the equation is the total field due to all charges, both inside and outside the surface.
• The law is most useful for calculating the electric field for charge distributions with a high degree of symmetry (spherical, cylindrical, planar).
• Gauss's law is a direct consequence of the inverse-square nature of Coulomb's law.

Given

• Electric field, $E_x = \alpha x^{1/2}$ with $\alpha = 800 \text{ N/C m}^{1/2}$
• $E_y = 0, E_z = 0$
• Side of cube, $a = 0.1$ m
• The cube is placed with one corner at the origin, extending from $x=a$ to $x=2a$.

To Find

(a) The net flux $\phi$ through the cube. (b) The net charge $q$ inside the cube.

Formula

• Flux, $\phi = \mathbf{E} \cdot \Delta\mathbf{S} = E \Delta S \cos\theta$
• Gauss's Law, $q = \phi \varepsilon_0$

Solution

(a) The electric field is only in the x-direction. Therefore, flux is non-zero only through the faces perpendicular to the x-axis (the left and right faces).

• Left Face: Located at $x=a$. The area vector points in the $-x$ direction, so $\theta = 180^\circ$. $E_L = \alpha a^{1/2}$ $\phi_L = E_L \cdot A = E_L a^2 \cos(180^\circ) = - \alpha a^{1/2} a^2 = -\alpha a^{5/2}$
• Right Face: Located at $x=2a$. The area vector points in the $+x$ direction, so $\theta = 0^\circ$. $E_R = \alpha (2a)^{1/2} = \alpha \sqrt{2} a^{1/2}$ $\phi_R = E_R \cdot A = E_R a^2 \cos(0^\circ) = (\alpha \sqrt{2} a^{1/2}) a^2 = \alpha \sqrt{2} a^{5/2}$ The net flux is the sum of the fluxes through all faces. $\phi_{net} = \phi_L + \phi_R = -\alpha a^{5/2} + \alpha \sqrt{2} a^{5/2} = \alpha a^{5/2}(\sqrt{2} - 1)$ Substitute the values: $\phi_{net} = 800 \times (0.1)^{5/2} (\sqrt{2} - 1) = 800 \times (0.00316) \times (0.414) \approx 1.05 \text{ N m}^2 \text{C}^{-1}$

(b) Using Gauss's Law to find the enclosed charge: $q = \phi_{net} \varepsilon_0 = (1.05 \text{ N m}^2 \text{C}^{-1}) \times (8.854 \times 10^{-12} \text{ C}^2 \text{N}^{-1} \text{m}^{-2})$ $q \approx 9.27 \times 10^{-12} \text{ C}$

Final Answer (a) The net flux is $1.05 \text{ N m}^2 \text{C}^{-1}$. (b) The charge within the cube is $9.27 \times 10^{-12} \text{ C}$.

Applications of Gauss's Law

Gauss's law simplifies electric field calculations for symmetric charge distributions.

Field due to an Infinitely Long Straight Uniformly Charged Wire

For a long wire with uniform linear charge density $\lambda$, the electric field is radial and perpendicular to the wire. We use a cylindrical Gaussian surface of radius r and length l.

• Flux passes only through the curved part of the cylinder.
• Flux $\phi = E \times (\text{Area}) = E \times (2\pi r l)$.
• Charge enclosed $q_{enc} = \lambda l$.
• Applying Gauss's law: $E(2\pi r l) = \frac{\lambda l}{\varepsilon_0}$.

The magnitude of the electric field is: $E = \frac{\lambda}{2\pi\varepsilon_0 r}$ The field strength decreases as $1/r$.

Field due to a Uniformly Charged Infinite Plane Sheet

For a large plane sheet with uniform surface charge density $\sigma$, the electric field is uniform and perpendicular to the sheet. We use a cylindrical or rectangular box as the Gaussian surface, piercing the sheet.

• Flux passes only through the two end faces (area A).
• Flux $\phi = EA + EA = 2EA$.
• Charge enclosed $q_{enc} = \sigma A$.
• Applying Gauss's law: $2EA = \frac{\sigma A}{\varepsilon_0}$.

The magnitude of the electric field is: $E = \frac{\sigma}{2\varepsilon_0}$ Remarkably, the field is constant and does not depend on the distance from the sheet.

Field due to a Uniformly Charged Thin Spherical Shell

For a thin spherical shell of radius R and total charge q (surface charge density $\sigma$), the field has spherical symmetry. We use a spherical Gaussian surface of radius r.

(i) Field outside the shell ($r > R$) The Gaussian surface encloses the entire charge q.

• Flux $\phi = E \times (4\pi r^2)$.
• Charge enclosed $q_{enc} = q$.
• Applying Gauss's law: $E(4\pi r^2) = \frac{q}{\varepsilon_0}$.

The electric field is: $\mathbf{E} = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \hat{\mathbf{r}}$ Outside the shell, the field is identical to that of a point charge q located at the center.

(ii) Field inside the shell ($r < R$) The Gaussian surface is inside the shell and encloses no charge.

• Charge enclosed $q_{enc} = 0$.
• Applying Gauss's law: $E(4\pi r^2) = 0$.

The electric field is: $E = 0$ The electric field is zero everywhere inside a uniformly charged thin spherical shell.