This segment introduces electric charges and fields, focusing on key concepts necessary for solving numerical problems. The chapter covers various topics including electrostatics, types of charges, and electric field intensity. Students are encouraged to take notes on important points to enhance understanding.

Electrostatics is defined as the study of properties of charges that are at rest. It examines how these charges behave and interact when stationary. There are two types of electric charges: positive and negative.

Charge is a property of matter that experiences a force when placed in an electromagnetic field. This definition is crucial for understanding the behavior of charges in different scenarios.

• ★Electrostatics is the study of properties of charges at rest.0:45
• ★There are two types of electric charges: positive and negative.1:30
• ★Charge experiences a force when placed in an electromagnetic field.2:15
• ★Understanding these concepts boosts confidence in solving problems.3:00

Electrostatics is the study of properties of charges that are at rest. A charge is a property of matter that experiences a force when placed in an electromagnetic field.

In any body, the number of protons equals the number of electrons, making it neutral. Therefore, all bodies in the universe are neutral due to this balance.

A body can become positively or negatively charged through the process of rubbing. When two bodies rub against each other, one becomes negatively charged and the other positively charged.

Examples of negatively charged bodies include plastic and silk cloth. Examples of positively charged bodies include glass rods and hair.

• ★Electrostatics is the study of charges at rest.4:59
• ★A charge experiences a force in an electromagnetic field.4:59
• ★All bodies in the universe are neutral due to equal protons and electrons.4:59
• ★Rubbing two bodies can create positive and negative charges.4:59
• ★Negatively charged bodies include plastic and silk cloth.4:59
• ★Positively charged bodies include glass rods and hair.4:59

Example: When a glass rod and an ebonite rod are rubbed together, the ebonite rod becomes negatively charged and the glass rod becomes positively charged.4:59

When a negatively charged body is brought near positively charged hairs, the hairs are attracted. This occurs because opposite charges attract each other.

A glass rod and an abon rod are initially neutral. When rubbed together, the glass rod becomes positively charged and the abon rod becomes negatively charged.

If two charged bodies come into contact, they can neutralize each other. For example, if the negatively charged abon rod touches a neutral object, it can become neutral.

When a charged body is in contact with the ground, it also becomes neutral. This is why charged bodies do not attract hairs when they are grounded.

• ★Hairs can act as a positively charged substance when in contact with a negatively charged body.10:00
• ★When a glass rod and an abon rod are rubbed together, the glass rod becomes positively charged and the abon rod becomes negatively charged.10:05
• ★Charged bodies can become neutral when they come into contact with each other.10:20
• ★When a charged body is in contact with the ground, it becomes neutral.10:35
• ★Certain materials like umber, plastic, ebonite, and silk consistently become negatively charged.11:20
• ★The mnemonic 'Apes' helps remember which materials become negatively charged: A for umber, P for plastic, E for ebonite, S for silk.11:30

Example: When the negatively charged abon rod touches a neutral object, both become neutral.10:40

When a positively charged body touches the ground, it becomes neutral by receiving electrons from the ground. For example, if there are 100 protons in the positively charged body, 100 electrons from the ground enter the body to neutralize it. Only electrons are transferable; protons do not move into the ground.

Conversely, when a negatively charged body touches the ground, it loses electrons to the ground and also becomes neutral. If there are 100 electrons in the negatively charged body, these 100 electrons will leave the body and enter the ground.

Electric field lines illustrate the behavior of charged bodies. For positively charged bodies, electric field lines point outward, while for negatively charged bodies, they point inward. Like charges repel each other, and unlike charges attract each other.

A charge at rest emits only an electric field. When a charge is in uniform motion, it emits both an electric field and a magnetic field.

• ★When a positively charged body touches the ground, it receives electrons and becomes neutral.14:56
• ★Electrons are the only transferable charge; protons do not move.14:56
• ★When a negatively charged body touches the ground, it loses electrons and becomes neutral.14:56
• ★Electric field lines for positive charges point outward; for negative charges, they point inward.14:56
• ★Like charges repel; unlike charges attract.14:56
• ★A charge at rest emits only an electric field; in motion, it emits both electric and magnetic fields.14:56

Example: If a positively charged body with 100 protons touches the ground, it receives 100 electrons to become neutral.14:56

Students often forget previous chapters if they do not revise regularly. Revising helps retain knowledge for exams like JEE Mains and NEET.

A charge at rest emits only an electric field. In uniform motion, it emits both electric and magnetic fields. In non-uniform motion, it emits electric fields, magnetic fields, and radiation.

To create a positively charged body, we must remove electrons. To create a negatively charged body, we add more electrons since electrons are transferable, while protons are not.

When two charged bodies come into contact, they share their charge equally. The formula for charge sharing is given by $Q_{share}=\frac{Q_1+Q_2}{2}$.

• ★Students forget earlier chapters without revision.19:56
• ★A charge at rest emits only an electric field.19:56
• ★In uniform motion, a charge emits electric and magnetic fields.19:56
• ★To create a positive charge, remove electrons.19:56
• ★To create a negative charge, add electrons.19:56
• ★Charge sharing formula: $Q_{share}=\frac{Q_1+Q_2}{2}$.19:56

Example: If one body has 100 protons and the other has 50 electrons, after contact, each body will have 25 protons.19:56

Different objects have varying amounts of charge based on their size and mass. For example, a pen has less charge than a book, and a screen has even more charge due to its larger size and mass. However, the charge inside each object is fixed and cannot be determined without calculations.

The quantization of charge states that every body has a fixed charge that remains constant. To calculate the total charge present in a body, use the formula $Q=\text{±}nA$, where $n$ represents the number of protons or electrons.

The net charge of any body is neutral, meaning it is zero because the number of protons equals the number of electrons. This is crucial: if asked for the net charge, you can simply write zero. However, if asked for the actual charge, you must use the formula to find it.

• ★Smaller objects have less charge than larger ones.24:55
• ★The formula for total charge is $Q=\text{±}nA$.25:05
• ★Net charge of a body is zero due to equal protons and electrons.25:20
• ★Charge quantization values ($n$) are integers, not fractions.25:50

Example: For a 10g pen, you can calculate the charge using the formula.25:10

A neutral body contains equal numbers of protons and electrons. For example, if there are 100 protons, there must also be 100 electrons. This balance keeps the body neutral.

To calculate the charge in 10 grams of water, you must determine the number of protons and electrons. The formula for total charge is $Q_{net}=+ne$ for protons and $Q_{net}=-ne$ for electrons.

The law of conservation of charge states that charge cannot be created or destroyed, only transferred. Only electrons can transfer between bodies, not protons.

When rubbing two different materials, such as a glass rod and a plastic rod, electrons may transfer from one to the other. For instance, if 100 electrons transfer from the glass rod to the plastic rod, the glass rod becomes positively charged.

• ★A neutral body has equal protons and electrons, e.g., 100 protons and 100 electrons.29:55
• ★Charge of an electron is negative, charge of a proton is positive.29:58
• ★Total charge in a body can be calculated using $Q_{net}=+ne$ for protons and $Q_{net}=-ne$ for electrons.30:15
• ★Charge cannot be created or destroyed; it can only be transferred.30:20
• ★Electrons are the only particles that transfer between bodies, not protons.30:25
• ★After rubbing, if 100 electrons transfer from glass to plastic, the glass becomes positively charged.30:30

Example: To find the charge in 10 grams of water, calculate the number of protons and electrons, resulting in $Q_{net}=+ne$ for protons and $Q_{net}=-ne$ for electrons.30:05

When a glass rod and a plastic rod are rubbed together, 100 electrons transfer from the glass rod to the plastic rod. This makes the plastic rod negatively charged, while the glass rod retains 100 protons, remaining neutral overall.

Electrons are the only particles that can be transferred between objects during rubbing. This process is known as friction, which is the method of charging by rubbing two objects together.

Conduction occurs when two charged bodies come into direct contact. The charge is shared according to the formula $\frac{Q_1+Q_2}{2}$. For example, if a charged body has 100 protons and it contacts an uncharged body, both will end up with 50 protons after contact.

Induction happens without direct contact. A charged body can influence the charge distribution in an uncharged body, leading to charge separation.

• ★100 electrons transfer from the glass rod to the plastic rod during rubbing.34:50
• ★Friction is the method of charging by rubbing two objects together.34:55
• ★Conduction involves direct contact between charged and uncharged bodies.35:05
• ★The formula for charge sharing during conduction is $\frac{Q_1+Q_2}{2}$.35:10
• ★Induction occurs without direct contact, influencing charge distribution.35:20

Example: If a charged body with 100 protons contacts an uncharged body, both end up with 50 protons after contact.35:15

An uncharged body can be charged without direct contact through induction. When two bodies are brought close together, protons and electrons interact, causing charge separation. For example, if there are 100 protons in an uncharged body, 100 electrons will crowd towards the positively charged area, leading to charge imbalance.

Grounding allows electrons to flow from the ground to the charged body. When a wire is connected to the ground, electrons from the ground enter the wire and move into the body, neutralizing the charge. This process balances the protons and electrons, making the body neutral.

After grounding, when the bodies are separated, the charge distributes evenly across the surface of the body. This results in one body becoming negatively charged while the other remains as a source charge with 100 protons.

Two charged bodies can be created simultaneously using the same induction process. When two uncharged bodies are in contact, bringing them close together causes charge separation. If they are then separated, one can become positively charged.

• ★An uncharged body can be charged through induction.39:50
• ★Electrons crowd towards positively charged areas, causing charge separation.40:05
• ★Grounding allows electrons to flow from the ground to neutralize charge.40:15
• ★Charge distributes evenly across the surface after separation.40:30
• ★Two charged bodies can be created simultaneously through induction.40:45
• ★Positive charges repel each other, causing protons to crowd on one side.40:50

Example: When two uncharged bodies are brought close, charge separation occurs, leading to one body becoming positively charged and the other negatively charged upon separation.40:55

A positively charged body can be created by having 100 protons and 100 electrons attracted to a source charge. When these charges are separated, the negative charge distributes uniformly on the sphere, resulting in a negatively charged body.

The inverse square law is crucial for understanding the forces between charged bodies. The force between two charges is never zero; it is always present in the universe. The forces acting on two charges, denoted as $F_1$ and $F_2$, represent their mutual interaction.

The main formula derived from the inverse square law states that the force $F$ is directly proportional to the product of the charges $Q_1$ and $Q_2$, and inversely proportional to the square of the distance $r$ between them. This can be expressed as $F\frac{1}{r^2}$.

The constant $K$ in the formula is given by $K=\frac{1}{4\text{π}{\text{ε}}_0}$, where ${\text{ε}}_0$ is the permittivity of free space. Remember, gravitational force does not depend on the medium, while Coulomb's force does.

• ★A positively charged body has 100 protons and 100 electrons.44:44
• ★The inverse square law is vital for electric force calculations.44:44
• ★Force is directly proportional to the product of charges and inversely proportional to the square of distance: $F\frac{Q_1{Q}_2}{r^2}$.44:44
• ★The constant $K=\frac{1}{4\text{π}{\text{ε}}_0}$ relates to the force between charges.44:44
• ★Coulomb's force depends on the medium; gravitational force does not.44:44
• ★Relative permittivity $K=\frac{{\text{ε}}_m}{{\text{ε}}_0}$.44:44

Example: When separating charges, a positively charged body and a negatively charged body are created simultaneously.44:44

Coulomb's law formula is given by $F=K\frac{Q_1{Q}_2}{R^2}$, where $K$ is the constant, $Q_1$ and $Q_2$ are the charges, and $R$ is the distance between them. In exam questions, the value of $K$ will be provided, and students must use it to calculate forces in free space.

When calculating forces between two spherical bodies or point charges, the distance $R$ is measured from the center of the spheres. Coulomb's force is not applicable for a rod with charge and a point charge; integration is required in such cases.

For quick calculations, remember the trick to solve numerical problems efficiently, often within 20-40 seconds. Common exam questions involve finding the net force on charges arranged in configurations like an equilateral triangle or a square.

Different teachers may explain these concepts in various ways, but my method focuses on efficiency. If you learn this approach, you can solve these types of numerical problems quickly.

• ★Coulomb's law is $F=K\frac{Q_1{Q}_2}{R^2}$.49:43
• ★Distance $R$ is measured from the center of spherical bodies.49:55
• ★Coulomb's force is not applicable for a rod and point charge; integration is needed.50:05
• ★Solve numerical problems in 20-40 seconds using the right tricks.50:20
• ★Common configurations include equilateral triangles and squares for charge arrangements.50:35

Example: Find the net force on three charges placed at the vertices of an equilateral triangle with side length $a$.50:45

The net force acting on a movable charge $q$ is influenced by fixed charges around it. If asked to find the net force on a charge, remember that this charge is movable while others are fixed.

The formula for calculating net force when two forces are present is $F_{net}={\text{F}}_1+{\text{F}}_2+{\text{F}}_3+...+{\text{F}}_n$. If the angle between the forces is $\theta$, the net force can be calculated using $$\begin{gathered} F_{net}= \\ sqrt{F}_1^2+F_2^2+2F_1{F}_2\text{cos}(\theta ) \end{gathered}$$.

For specific angles, if $\theta =0^{\text{o}}$, then $F_{net}=2F$. If $\theta =30^{\text{o}}$, then $$\begin{gathered} F_{net}= \\ sqrt3F \end{gathered}$$. If $\theta =45^{\text{o}}$, then $$\begin{gathered} F_{net}= \\ sqrt2F \end{gathered}$$. If $\theta =60^{\text{o}}$, then $$\begin{gathered} F_{net}= \\ sqrt3F \end{gathered}$$. If $\theta =90^{\text{o}}$, then $F_{net}=0$ if forces are equal.

Electric field intensity is defined as $E=K\frac{Q}{r^2}$, where $K=\frac{1}{4\pi {ϵ}_0}$. In electric field intensity problems, the charge is not present at the point of interest.

• ★The net force on a movable charge is influenced by fixed charges.54:41
• ★Use $F_{net}=\sqrt{F_1^2+F_2^2+2F_1{F}_2\text{cos}(\theta )}$ for net force with angles.54:41
• ★If $\theta =0^{\text{o}}$, then $F_{net}=2F$; if $\theta =30^{\text{o}}$, then $F_{net}=\sqrt{3}F$.54:41
• ★Electric field intensity is $E=K\frac{Q}{r^2}$.54:41
• ★In electric field intensity problems, the charge is not present at the point of interest.54:41

Example: If two equal forces act at an angle of $30^{\text{o}}$, then $F_{net}=\sqrt{3}F$.54:41

The net electric field acting on a charge $q$ is represented by capital $E$. For an arc with charge $Q$ and radius $R$, we can calculate the electric field intensity using specific formulas.

The formula for electric field intensity along the x-axis is $E_{X-axis}=\frac{k\lambda }{r}{\int }_{{\theta }_1}^{{\theta }_2}\cos (\theta )d(\theta )$. For the y-axis, it is $E_{Y-axis}=\frac{k\lambda }{r}{\int }_{{\theta }_1}^{{\theta }_2}\sin (\theta )d(\theta )$.

Linear charge density $\lambda$ is defined as total charge divided by total length. Substitute given angle values into these formulas to find the electric field intensity.

The perpendicular distance from the charge to the point of interest is critical for solving problems. If this distance is not correctly identified, the numerical results will be incorrect.

• ★The net electric field on charge $q$ is $E$.59:40
• ★Electric field intensity along the x-axis: $E_{X-axis}=\frac{k\lambda }{r}{\int }_{{\theta }_1}^{{\theta }_2}\cos (\theta )d(\theta )$59:50
• ★Electric field intensity along the y-axis: $E_{Y-axis}=\frac{k\lambda }{r}{\int }_{{\theta }_1}^{{\theta }_2}\sin (\theta )d(\theta )$59:55
• ★Linear charge density $\lambda =\frac{Q}{L}$, where $Q$ is total charge and $L$ is total length.60:05
• ★Perpendicular distance from charge to point is essential for calculations.60:20

Example: For angles from $30^{\circ }$ to $60^{\circ }$, substitute these values into the formulas to find the electric field intensity.60:10

To calculate electric field intensity from finite and infinite charged wires, we use angles alpha and beta. For an infinite wire, both angles are 90° when determining electric field intensity.

The formula for electric field intensity due to an infinite wire is derived as $E_{x-axis}=\frac{2k\lambda }{r}$, where $k$ is a constant and $\lambda$ is the linear charge density. The y-axis electric field intensity is zero.

Linear charge density $\lambda$ is defined as total charge per unit length. For two-dimensional distributions, area charge density $\sigma$ is defined as total charge per unit area. For three-dimensional distributions, volume charge density $\rho$ is defined as total charge per unit volume.

Electric field intensity due to an infinite charge sheet is given by $E=\frac{\sigma }{2{ϵ}_0}$. This intensity remains constant regardless of the distance from the sheet.

• ★For an infinite wire, angles alpha and beta are both 90°.64:38
• ★The electric field intensity formula for an infinite wire is $E_{x-axis}=\frac{2k\lambda }{r}$.64:38
• ★Linear charge density $\lambda$ is total charge per unit length.64:38
• ★Area charge density $\sigma$ is total charge per unit area.64:38
• ★Volume charge density $\rho$ is total charge per unit volume.64:38
• ★Electric field intensity due to an infinite charge sheet is $E=\frac{\sigma }{2{ϵ}_0}$.64:38
• ★Electric field intensity from an infinite sheet remains constant with distance.64:38

To calculate the electric field intensity on the axis of a circular charge distribution, use the formula $E=\frac{KQ\cdot r}{(R^2+r^2{)}^{3/2}}$. Here, $r$ is the distance from the center, and $R$ is the radius of the circle.

The maximum electric field intensity occurs at a distance of $\frac{R}{2}$ from the center of the circular charge. The maximum value is given by $E_{max}=\frac{2}{3}\cdot \frac{KQ}{R^2}$.

An electron placed on the axis of the circular charge will perform simple harmonic motion. Initially, its velocity is zero, but as it moves towards the charge, its speed increases.

The electric field intensity at the center of the circular charge is zero. For an arc of the charge, the electric field can be calculated using the integration formula $E=K\lambda \int \frac{sin\Theta }{R}d\Theta$.

• ★The formula for electric field intensity on the axis is $E=\frac{KQ\cdot r}{(R^2+r^2{)}^{3/2}}$69:37
• ★Maximum electric field intensity occurs at $\frac{R}{2}$ with a value of $E_{max}=\frac{2}{3}\cdot \frac{KQ}{R^2}$69:45
• ★Electric field intensity at the center of the circular charge is zero69:55
• ★An electron on the axis performs simple harmonic motion due to the electric field70:05
• ★Integration formula for electric field due to an arc is $E=K\lambda \int \frac{sin\Theta }{R}d\Theta$70:15
• ★For a $90^{\circ }$ angle, electric field intensity is $E=\sqrt{2}\cdot \frac{K\lambda }{R}$70:25
• ★Net electric field from two segments at $90^{\circ }$ is $E_{net}=2\cdot \frac{K\lambda }{R}$70:35

Example: An electron on the axis starts with zero velocity and accelerates towards the positive charge due to the electric field.70:10

The electric field intensity due to an arc of 180° is given by the formula $E=\frac{2K\lambda }{R}$. This represents the electric field at the center of the arc.

When calculating electric field intensity at the center of a circular charge distribution, components can cancel each other out. For example, in a circular arrangement, the electric field intensity can be zero due to symmetry.

Inertia affects the motion of an electron in an electric field. As it moves towards the center, the electric field intensity becomes zero, but the electron continues moving forward due to inertia, performing simple harmonic motion (SHM).

Conducting spheres have charge only on the surface, while non-conducting spheres have charge distributed both inside and outside. The electric field intensity outside a conducting shell is given by $E_{out}=\frac{kQ}{r^2}$, where $Q$ is the charge and $r$ is the distance from the center.

• ★Electric field intensity due to an arc of 180° is $E=\frac{2K\lambda }{R}$.74:36
• ★Components of electric field intensity can cancel each other out, leading to a net intensity of zero.74:45
• ★Inertia causes an electron to perform simple harmonic motion (SHM) in an electric field.74:55
• ★Conducting spheres have charge only on the surface; non-conducting spheres have charge throughout.75:05
• ★Electric field intensity outside a conducting shell is $E_{out}=\frac{kQ}{r^2}$.75:15

Example: An electron moving in an electric field experiences maximum velocity at certain points, demonstrating SHM.74:40

The electric field intensity inside a charged shell is zero. For a charged sphere, the electric field intensity inside is given by $E_{in}=\frac{kQr}{R^3}$. At the center of a charged sphere, the electric field intensity is also zero.

The formula $E=\frac{kQ}{r^2}$ applies to both point charges and spherical charged bodies. Graphs of electric field intensity versus radius differ for conducting and non-conducting bodies. Outside a charged body, the electric field intensity is inversely proportional to $R^2$.

Inside a non-conducting sphere, the electric field intensity is directly proportional to the radius, represented as a straight line graph. The maximum electric field intensity occurs at the surface of charged bodies.

• ★Electric field intensity inside a charged shell is zero.79:34
• ★Electric field intensity inside a charged sphere is $E_{in}=\frac{kQr}{R^3}$.79:45
• ★At the center of a charged sphere, electric field intensity is zero.79:55
• ★Outside a charged body, electric field intensity is inversely proportional to $R^2$.80:05
• ★Inside a non-conducting sphere, electric field intensity is directly proportional to the radius.80:15
• ★Maximum electric field intensity occurs on the surface of charged bodies.80:25

Example: The electric field intensity at the center of a charged sphere is $E_{centre}=0$.79:55

The electric field intensity inside a hollow infinite cylinder is zero. On the surface, it is non-zero, while outside it is inversely proportional to the distance from the center.

Graphs are key for understanding electric field intensity. You need to identify which graph corresponds to which scenario rather than memorizing formulas.

When a charge moves in an electric field, you need to calculate its velocity and angle as it exits. The minimum velocity required for an electron to exit the field can be determined from the conditions given.

An electric dipole consists of two equal but opposite charges separated by a distance of 2L. The dipole moment, denoted as $p$, is given by $p=Q\times 2L$ and is a vector quantity directed from the negative charge to the positive charge.

• ★Electric field intensity inside a hollow infinite cylinder is zero.84:32
• ★Graphs are more important than formulas for electric field intensity.84:35
• ★Minimum velocity for an electron to exit the field can be calculated.84:40
• ★An electric dipole consists of two equal but opposite charges.84:45
• ★Dipole moment is given by $p=Q\times 2L$.84:50
• ★Direction of dipole moment is from negative charge to positive charge.84:55

Example: The angle at which the electron exits the field is given by $\theta ={\tan }^{-1}\left(\frac{E_{e}l}{m_e{v}_0}\right)$ and its velocity by $V=\sqrt{v_0^2+{\left(\frac{E_{e}l}{m_e{v}_0}\right)}^2}$.84:45

The net electric field intensity and the dipole moment direction are aligned, with an angle $\theta$ between them. The dipole moment is defined as $P=Q\times 2L$, where $Q$ is the charge and $L$ is the length of the dipole.

The electric field intensity due to a dipole along the equatorial line is given by $E=k\frac{p}{r^3}$ for small dipole moments. Here, $r$ is measured from the origin.

When calculating the electric field intensity at any point in space, the formula is $E=k\frac{p}{r^3}(3\text{cos}\theta +1)$. The angle $\theta$ can vary, affecting the result.

The torque experienced by a dipole in an electric field is given by the equation $\tau =\text{P}\times \text{E}$. The right-hand thumb rule helps determine the direction of torque.

• ★The dipole moment is $P=Q\times 2L$.89:30
• ★Electric field intensity along the equatorial line is $E=k\frac{p}{r^3}$.89:40
• ★The formula for electric field intensity at any point is $E=k\frac{p}{r^3}(3\text{cos}\theta +1)$.90:00
• ★Torque is calculated as $\tau =\text{P}\times \text{E}$.90:10
• ★Use the right-hand thumb rule for torque direction.90:20

Example: If a dipole is placed in an electric field at an angle $\theta$, it experiences torque calculated by $\tau =\text{P}\times \text{E}$.90:25

For an electric dipole in an electric field, if $\theta =90^{\text{o}}$, the torque is given by $\tau =PE\text{sin}\theta$. At $\theta =0^{\text{o}}$, it is in stable equilibrium, while at $\theta =180^{\text{o}}$, it is in unstable equilibrium. The potential energy of the dipole is stored due to its orientation, expressed as $U=-PE\text{cos}(\theta )$.

Electric flux, denoted by $\text{Φ}$, is calculated using the formula $\text{Φ}=EA\text{cos}\theta$, where $E$ is the electric field and $A$ is the area. Electric flux is a scalar quantity measured in volts per meter. For closed surfaces, the net electric flux is zero when the flux entering equals the flux exiting.

According to Gauss's Law, the total electric flux from a closed surface is proportional to the charge enclosed, expressed as ${\text{Φ}}_E=\frac{q_{inside}}{{\text{ε}}_0}$. This relationship is critical for solving problems in exams like NEET and JEE Mains.

• ★Torque for electric dipole: $\tau =PE\text{sin}\theta$94:28
• ★Stable equilibrium at $\theta =0^{\text{o}}$; unstable at $\theta =180^{\text{o}}$.94:28
• ★Potential energy of dipole: $U=-PE\text{cos}(\theta )$94:28
• ★Electric flux formula: $\text{Φ}=EA\text{cos}\theta$94:28
• ★Net electric flux for closed surfaces is zero when entering equals exiting.94:28
• ★Total electric flux: ${\text{Φ}}_E=\frac{q_{inside}}{{\text{ε}}_0}$94:28

To find electric field intensity using Gauss's law, a closed loop is necessary. This closed surface is called a Gaussian surface. For a point charge $Q$, the electric field intensity $E$ can be derived using the formula $E=\frac{Q}{4\pi {ϵ}_0{r}^2}$, where $r$ is the distance from the charge.

The relationship between electric field and charge can be expressed as ${\Phi }_E=\int EdA=\frac{q_{inside}}{{ϵ}_0}$. Practicing numerical problems related to electric fields is essential for mastering this topic.

• ★A closed loop is required to apply Gauss's law.99:28
• ★Gaussian surface is the closed surface used in Gauss's law.99:35
• ★Electric field intensity is given by $E=\frac{Q}{4\pi {ϵ}_0{r}^2}$.99:45
• ★Regular revision of key concepts is crucial for exam success.100:10

Example: For a charge $Q$ at distance $r$, the electric field intensity is $E=\frac{Q}{4\pi {ϵ}_0{r}^2}$.100:00