1. Electric Charge:

The fundamental property of any substance which produces electric and magnetic fields.

(i) Charge cannot exist without mass, but mass can exist without charge.

(ii) Charge is independent of its velocity.

(iii) Charge at rest produces electrostatic field.

(iv) Charge in motion produces both electric and magnetic fields.

(v) Accelerating charge produces electromagnetic waves.

(vi) Charge of a body is always equal to the integral multiple of charge of an electron, i.e., $q = \pm n e$ .

(vii) Total charge of a system is always equal to the algebraic sum of the individual particles of the system.

(viii) Repulsion is a sure test for electrification.

(ix) When a neutral body loses electrons, it gains a positive charge and its mass decreases.

(x) When a neutral body gains electrons, it becomes negatively charged and its mass increases.

(xi) When a charged body is brought close to the neutral body, then charges induce (polarize) in the neutral body.

(xii) If two charged bodies are connected by a conducting wire, then charge transfer takes place until their potentials become the same.

(xiii) If two metal spheres of radii $r_{1}$ and $r_{2}$ having charge $q_{1}$ and $q_{2}$ are connected by a metal wire, then final charges on the spheres are,

$q_{1}^{'} = (\frac{q_{1} + q_{2}}{r_{1} + r_{2}}) r_{1}$ and $q_{2}^{'} = (\frac{q_{1} + q_{2}}{r_{1} + r_{2}}) r_{2}$

(xiv) Charge on a conductor:

The charge given to a conductor always resides on its outer surface. This is why a solid and hollow conducting sphere of the same outer radius will hold a maximum equal charge. If the surface is uniform, then the charge distributes uniformly on the surface and for irregular surfaces, the distribution of charge and charge density is not uniform. It is maximum where the radius of curvature is minimum and vice versa, i.e., $σ \propto$ $(\frac{1}{R})$ . This is why the charge leaks from sharp points.

2. Charge Distribution:

It may be of two types:

(i) Discrete distribution of charge:

A system consisting of ultimate individual charges.

(ii) Continuous distribution of charge:

An amount of charge distributes uniformly or non-uniformly on a body. It is of the following three types:

(a) Line charge distribution:

Charge on a line, e.g., charged straight wire, circular charged ring, etc.

Linear charge density, $λ = \frac{Charge}{Length}$ .

S.I. unit is $C m^{- 1}$ .

Dimension: $L^{- 1} TA$ .

(b) Surface charge distribution:

Charge distributed on a surface, e.g., a plane sheet of charge, conducting sphere, conducting cylinder.

Surface charge density, $σ = \frac{Charge}{Area}$

S.I. unit is $C m^{- 2}$ .

Dimension: $L^{- 2} TA$ .

(c) Volume charge density: Charge distributes throughout the volume of the body, e.g., charge on a dielectric sphere, etc.

Volume charge density, $ρ = \frac{Charge}{Volume}$

S.I. unit is $C m^{- 3}$

Dimension: $L^{- 3} TA$

3. Coulomb's Law:

The electrostatic force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them.

(i) Consider two points charges $q_{1}$ and $q_{2}$ are separated by a distance $r$ in a vacuum, then the electrostatic force between them is

$F_{vac} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$ , where $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} C^{- 2}$ .

(ii) If the charges $q_{1}$ and $q_{2}$ are separated by a distance $r$ in a medium of dielectric constant $K$ , then force between them is

$F_{med} = \frac{1}{4 π ε_{0} K} \frac{q_{1} q_{2}}{r^{2}} = \frac{1}{4 π ε_{0} ε_{r}} \frac{q_{1} q_{2}}{r^{2}}$

$= \frac{1}{4 π ε} \frac{q_{1} q_{2}}{r^{2}}$

(iii) $F_{med} = \frac{F_{vac}}{K}$

(iv) Permittivity $(ε)$ : The ability of the medium to allow electric field lines to pass through it is called permittivity.

(v) Relative permittivity $(ε_{r})$ (or) dielectric constant $(K)$ :

(a) The ratio of the permittivity of a medium to the permittivity of vacuum.

(b) $ε_{r} = \frac{ε}{ε_{0}} = \frac{F_{vac}}{F_{med}}$

(d) $ε_{r} = 1$ for vacuum and approximately equal to $1$ for air.

(e) $ε_{r} \geq 1$ always.

(f) $ε_{r}$ increases with an increase in the conductivity of the medium.

(g) $ε_{r} = \infty$ for a perfect conductor.

(vi) Coulomb’s law in vector form:

(a) Force on $q_{2}$ by $q_{1}$ is,

${\vec{F}}_{21} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{3}} \vec{r}$

(b) Charges in a plane (or) space:

Force on $q_{2}$ by $q_{1}$ is ${\vec{F}}_{21} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{{|\vec{r_{2}} - \vec{r_{1}}|}^{3}} (\vec{r_{2}} - \vec{r_{1}})$

(vii) Coulomb’s law is applicable to the static point charges.

(viii) The electrostatic force between two charges is independent of the pressure (or) absence of any other charges.

(ix) Electrostatic force is conservative.

(x) The distance between two charges in vacuum is $r$ , when the force between the same charges in a medium of dielectric constant $K$ to experience the same force is $r' = \frac{r}{\sqrt{K}}$ .

(xi) $ρ =$ density of ball, $σ =$ density of the liquid.

To remain $θ$ same in both cases, the dielectric constant of the liquid is $K = \frac{ρ}{ρ - σ}$ .

(xii) Two particles each of charge $q$ and with masses $m_{1}$ and $m_{2}$ are suspended from a common point with the help of identical strings as shown, then

(a) $α = β$ , if $m_{1} = m_{2}$

(b) $α > β$ , if $m_{1} < m_{2}$

(xiii) A thin wire ring of radius $r$ has a charge $q$ . When a point charge $q_{0}$ is placed at its centre, then the tension developed in the ring is

$T = \frac{q q_{0}}{8 π^{2} ε_{0} r^{2}}$

(xiv) If two like charges $q_{1}$ and $q_{2}$ are separated by a distance $d$ , then one null point exists in between them.

The distance of a null point from the charge of smaller magnitude is $x = \frac{d}{1 + \sqrt{\frac{q_{2}}{q_{1}}}}$ .

(xv) If two unlike charges are separated, then one null point exists outside the charges and near to the smaller charge. Let $x$ be the distance from the charge of smaller magnitude, then

$\frac{q_{1}}{{(d + x)}^{2}} = \frac{q_{2}}{x^{2}}$

(xvi) Three identical particles each of charge $q$ are placed at the corners of an equilateral triangle of side $a$ . Then,

Force on any one charge is $F = \frac{\sqrt{3} k q^{2}}{a^{2}}$ .

(xvii) Four identical particles each of charge $q$ are placed at the corners of a square of side $a$ . Then,

Force on any one charge is $F = (\frac{1 + 2 \sqrt{2}}{2}) \frac{k q^{2}}{a^{2}}$ .

(xviii) Two identical particles each of charge $q$ are moving in a circle of radius $r$ under their mutual force of attraction. Then,

Angular velocity of each particle is $ω = \sqrt{\frac{k q^{2}}{4 m r^{3}}}$ or $ω = \sqrt{\frac{q^{2}}{16 π ε_{0} m r^{3}}}$ .

4. Electric Field:

The space around a charged particle in which other particles can be influenced.

5. Electric Field Intensity $(E)$ :

The force experienced by a unit positive charge at a point in the electric field.

(i) $\vec{E} = \frac{\vec{F}}{q}$

(ii) It is a vector.

(iii) S.I. unit: $N C^{- 1}$ and dimension: ${MLT}^{- 1} A$ .

(iv) The direction of $\vec{E}$ is away from positive charge and towards negative charge.

(v) Electric field due to a point charge:

$\vec{E} = \frac{1}{4 π ε_{0}} \frac{q}{r^{3}} \vec{r}$

(vi) The force experienced by a charge $q$ in an electric field of intensity $E$ is $\vec{F} = q \vec{E}$ .

If the charge is positive, then the force is directed in the direction of the field while if the charge is negative force acts on it in the opposite direction of the field.

6. Electric Field Due to a Uniformly Charged Ring:

(i) Electric field on the axis of a ring:
Consider a circular ring of radius $R$ with a charge $Q$ distributed uniformly over its length. Then, the electric field intensity at a distance $x$ from the centre on its axis is

${\vec{E}}_{axis} = \frac{1}{4 π ε_{0}} \frac{Q x}{{(x^{2} + R^{2})}^{3 / 2}} \hat{i}$

(ii) Electric field at the centre of the ring is $E_{centre} = 0$ .

(iii) Electric field is maximum at $x = \pm \frac{R}{\sqrt{2}}$ .

(iv) The maximum value of the electric field is $E_{m a x} = \frac{Q}{6 \sqrt{3} π ε_{0} R^{2}}$ .

7. Electric Field Due to a Finite Line Charge of Linear Density $λ$ :

(i) ${\vec{E}}_{x} = \frac{λ}{4 π ε_{0} r} (\sin α + \sin β) \hat{i}$ , where $λ$ is linear charge density and $r$ is the perpendicular distance.

(ii) ${\vec{E}}_{y} = \frac{λ}{4 π ε_{0} r} (\cos β - \cos α) (- \hat{j})$

(iii) $E_{net} = \sqrt{E_{x}^{2} + E_{y}^{2}}$ and $tan θ = \frac{E_{y}}{E_{x}}$

(iv) Field due to infinite line charge:

(a) $α = β = 90 °$

(b) $E_{x} = \frac{λ}{4 π ε_{0} r} (s i n 90 ° + s i n 90 °) = \frac{λ}{2 π ε_{0} r}$

(d) ${\vec{E}}_{net} = {\vec{E}}_{x} = \frac{λ}{2 π ε_{0} r} \hat{i}$

(v) Field due to semi-infinite line charge:

(a) $α = 90 °; β = 0 °$

(b) $E_{x} = \frac{λ}{4 π ε_{0} r} (\sin 90 ° + \sin 0 °) = \frac{λ}{4 π ε_{0} r} \hat{i}$

(d) $E_{net} = \sqrt{2} E_{x} = \frac{λ}{2 \sqrt{2} π ε_{0} r}$ and $θ = 45 °$

8. Electric Field Due to a Uniformly Charged Disc:

$σ =$ surface charge density,

$R =$ radius of the disc.

(i) Electric field on the own axis of the disc is $\vec{E} = \frac{σ}{2 ε_{0}} (1 - \frac{x}{\sqrt{x^{2} + R^{2}}}) \hat{i}$ , where $x$ is the distance from the centre on the axis.

(ii) For $R ≫ x,$ we get $E = \frac{σ}{2 ε_{0}}$

(iii) Electric field due to an infinite disc is independent of distance from it.

9. Electric Field Due to a Finite Charged Wire on Its Axis:

$λ =$ linear charge density, $l =$ length of the wire and $a =$ distance of a point at distance $a$ from one end of the wire.

(i) Electric field: $E = \frac{λ}{4 π ε_{0}} (\frac{1}{a} - \frac{1}{l + a})$ .

(ii) If $a > > l$ , then $E = \frac{λ l}{4 π ε_{0} a^{2}} = \frac{Q}{4 π ε_{0} a^{2}}$ .

(iii) The rod behaves as a point charge for $a > > l$ .

10. Electric Field Due to a Charged Arc:

$λ =$ linear charge density, $r =$ radius of the ring and $ϕ =$ the angle subtended by the arc at its centre.

(i) $E = \frac{λ}{2 π ε_{0} r} s i n (\frac{ϕ}{2})$ , where $ϕ$ is in $radian$ .

(ii) For semi-circular ring, $ϕ = π$ . So, the electric field is $E = \frac{λ}{2 π ε_{0} r} = \frac{Q}{2 π^{2} ε_{0} r^{2}}$ .

(iii) For a quarter circle, $ϕ = \frac{π}{2}$ and $E = \frac{λ}{2 \sqrt{2} π ε_{0} r}$ .

11. Electric field at the centre of a charged hemispherical shell is $E = \frac{σ}{4 ε_{0}}$ , where $σ$ is surface charge density.

12. Electric Field Due to a Spherical Shell (or) Metal Sphere:

Consider a thin spherical shell of radius $R$ with a charge $Q$ distributed over its surface uniformly.
(i) Field inside the shell, $E_{in} = 0$ .

(ii) Field on the surface of the shell, $E_{sur} = \frac{1}{4 π ε_{0}} \frac{Q}{R^{2}}$ .

(or) $E_{sur} = \frac{σ}{ε_{0}}$ , where $Q$ is the charge on the shell, $R$ is the radius of shell and $σ$ is surface charge density.

(iii) Field at a radial distance $x$ , outside the shell is

$E_{out} = \frac{1}{4 π ε_{0}} \frac{Q}{x^{2}} (x > R)$ (or) $E_{out} = \frac{σ R^{2}}{ε_{0} x^{2}}$

13. Electric Field Due to a Uniformly Charged Non-conducting Sphere:

(i) Field on the surface, $E_{sur} = \frac{Q}{4 π ε_{0} R^{2}}$ (or) $E_{sur} = \frac{ρ R^{2}}{3 ε_{0}}$ , where $Q =$ charge of the sphere, $R =$ radius of the sphere and $ρ =$ volume charge density.

(ii) Field at a radial distance $x$ , outside the sphere is

$E_{out} = \frac{Q}{4 π ε_{0} x^{2}} (x > R)$ (or) $E_{out} = \frac{ρ R^{3}}{3 ε_{0} x^{2}}$

(iii) Field at a distance $r$ from the centre and inside the sphere is,

$E_{inside} = \frac{Q x}{4 π ε_{0} R^{3}}$ (or) $E_{inside} = \frac{ρ x}{3 ε_{0}} (for x \leq R)$

(iv) Field at the centre, $E_{c e n t r e} = 0$ .

14. Electric Field Due to a Long Uniformly Charged Cylindrical Shell (or) Conducting Cylinder:

(i) Electric field inside the shell, $E_{in} = 0$ (for $x < R$ ).

(ii) Field outside the shell, $E_{out} = \frac{σ R}{ε_{0} x}$ , where $σ =$ surface charge density and $x =$ radial distance.

(iii) Field on the surface, $E_{sur} = \frac{σ}{ε_{0}}$ .

15. Electric Field Due to a Long Uniformly Charged Non-conducting cylinder:

Consider a non-conducting cylinder of volume charge density $ρ$ and radius $R$ .

(i) Electric field inside the cylinder, $E_{in} = \frac{ρ x}{2 ε_{0}} (x < R)$ , where $ρ =$ volume charge density and $x =$ radial distance.

(ii) Electric field outside the cylinder, $E_{out} = \frac{ρ R^{2}}{2 ε_{0} x}$ (for $x > R$ ).

(iii) Electric field on the surface, $E_{sur} = \frac{ρ R}{2 ε_{0}}$ .

16. Electric Field Due to a Sphere With Cavity:

(i) $\vec{E} = {\vec{E}}_{actual} - {\vec{E}}_{cavity}$

(ii) Electric field inside the cavity:

(a) $\vec{E} = \frac{ρ \vec{a}}{3 ε_{0}}$ , where $ρ =$ volume charge density and $\vec{a} =$ position vector of the centre of the cavity relative to the centre of the sphere.

(b) Field inside the cavity is uniform and its direction is parallel to radial vector $\vec{a}$ .

(iii) Electric field outside the cavity but inside the sphere:
$\vec{E} = \frac{ρ}{3 ε_{0}} [\vec{r} - {(\frac{b}{|\vec{r} - \vec{a}|})}^{3} (\vec{r} - \vec{a})]$

$a =$ distance between centres of the cavity and the actual sphere,

$b =$ radius of the cavity,

$R =$ radius of sphere and $r =$ radial distance of a point inside the sphere.

(iv) Electric field outside the sphere:

$\vec{E} = \frac{ρ}{3 ε_{0}} [{(\frac{R}{r})}^{3} \vec{r} - {(\frac{b}{|\vec{r} - \vec{a}|})}^{3} (\vec{r} - \vec{a})]$

17. Electric Field Due to Cylinder With Cavity and Inside the Cavity:

Consider a non-conducting cylinder with volume charge density $ρ$ has a cylindrical cavity whose axis is at a radial distance $a$ from the axis of the actual cylinder. The electric field inside the cavity is,
$\vec{E} = \frac{ρ \vec{a}}{2 ε_{0}}$

18. Electrostatic Pressure on a Charged Conductor:

(i) $\frac{d F}{d A} = P = \frac{σ^{2}}{2 ε_{0}} = \frac{1}{2} ε_{0} E^{2}$ , where $σ =$ surface charge density and $E =$ electric field on the surface.

(ii) Force on part of the conductor is $d F = P_{e l e c t r o s t a t i c} \times d A$ , where $d A =$ surface area of the part of the conductor.

(iii) Electrostatic stress, $P \propto σ^{2}$ (or) $P \propto E^{2}$ , which can make the charge fly apart but held by the mechanical force of the conductor.

19. Electrostatic force of interaction between two halves of a spherical conductor of radius $R$ carrying a charge $Q$ is $F = \frac{Q^{2}}{32 π ϵ_{0} R}$ .

20. Motion of a Charged Particle in a Uniform Electric Field:

(i) When a charged particle is released in a uniform electric field from rest:

Consider a particle of charge $Q$ and mass $m$ is released from rest in an electric field of strength $E$ . The particle will experience an electric force that causes its motion.

(a) The force experienced by the charged particle is $F = Q E$ .

(b) Acceleration produced by this force is $a = \frac{F}{m} = \frac{Q E}{m}$ .

(c) Consider a charged particle is released at point $A$ at time $t = 0$ , and it reaches the point $B$ after time $t$ where its velocity becomes $v$ .
Take $V =$ difference between $A$ and $B$ , separation between $A$ and $B$ as $S$ .

$v_{B} = \frac{Q E}{m} t$ $= \sqrt{\frac{2 Q Δ V}{m}} = \sqrt{2 \frac{Q E}{m} S}$

(d) Kinetic energy gained by the particle in time $t$ is,

$K = \frac{1}{2} m v^{2} = \frac{1}{2} m {(\frac{Q E t}{m})}^{2} = \frac{Q^{2} E^{2} t^{2}}{2 m}$

or $K = \frac{1}{2} m \times \frac{2 Q V}{m} = Q V$

(ii) According to work-energy theorem, we can say that gain in kinetic energy $=$ work done in the displacement of charge, i.e., $W = Q V$ , where $V =$ potential difference between the two positions of charge $Q$ .

(iii) Work done in displacing a charge in an electrostatic field is path independent.

Work done from $A$ to $B$ is same along all the paths $I$ , $II$ and $III$ shown in the below figure.

(iv) When a charged particle enters with an initial velocity at a right angle to the uniform field:

When a charged particle enters perpendicularly in an electric field, it describes a parabolic path as shown.

(a) Equation of trajectory: Throughout the motion, the particle has uniform velocity along the $X$ -axis and horizontal displacement $(x)$ is given by the equation $x = u t$ .

(b) The motion of the particle is accelerated along $Y$ -axis.

So, $y = \frac{1}{2} (\frac{Q E}{m}) {(\frac{x}{u})}^{2}$ ; this is the equation of parabola which shows $y \propto x^{2}$ .

So, $v = | \vec{v} | = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{u^{2} + \frac{Q^{2} E^{2} t^{2}}{m^{2}}}$

(d) If $β$ is the angle made by $v$ with $X$ -axis at time $t$ , then

$\tan β = \frac{v_{y}}{v_{x}} = \frac{Q E t}{m u}$

21. Electric Field Lines:

(i) Definition: An imaginary line in the field whose tangent represents the direction of the field at a point is called an electric field line.

(ii) Properties of electric field lines:

(a) Electric field lines originate at positive charge and terminate at negative charge.

(b) Tangent to the field line at any point gives the direction of the field at that point.

(c) Field lines never intersect each other. If they do, then the field should have two directions at that point that is impossible.

(d) Field lines are always normal to conducting surface.

(e) Field lines do not exist inside a conductor.

(f) The electrostatic field lines are open loops (while magneto static field lines form closed loops).

Note: Induced electric field lines are closed loops.

(g) The number of lines originating or terminating on a charge is proportional to the magnitude of charge, i.e., $|Q| \propto$ number of lines. In the following figure $|Q_{A}| > |Q_{B}|$ .

(h) If the lines of forces are equidistant and parallel straight lines of the field are uniform, and if either the line of force is not an equidistant or straight line or both, then the field will be non-uniform; also, the density of the field lines is proportional to the strength of the electric field.

22. Electric Flux $(ϕ)$ :

(i) Electric flux leaving the surface is given by $ϕ = E A \cos θ$ , if the field is uniform at every point on the surface. Here, $θ$ is the angle between the electric field $(\vec{E})$ and the area vector $(\vec{A})$ .
(ii) $ϕ = \int_{s u r f a c e}^{} \vec{E} \cdot d \vec{A}$ , if the field is non-uniform.

(iii) Leaving flux is $+ ve$ and entering flux is $- ve$

(iv) Net flux passing through a surface is always zero.

23. Gauss’s Law:

The net electric flux leaving a closed surface is always equal to $\frac{1}{ε_{0}}$ times the net charge enclosed.

$\oint_{s} \vec{E} \cdot d \vec{s} = \frac{1}{ε_{0}} q_{enclosed}$

(i) Electric flux through a circular disc $(φ)$ :

A point charge $q$ is placed at a distance $b$ from the centre of the circular disc of radius $R$ is, $φ = \frac{q}{2 ε_{0}} (1 - \frac{b}{\sqrt{b^{2} + R^{2}}}) = \frac{q}{2 ε_{0}} (1 - \cos ϕ)$

(ii) Electric flux through the lateral surface of a cylinder due to a point charge at its centre:
Consider a point charge $q$ is placed at the centre of the cylinder of length $l$ and radius $R$ . The flux through the curved surface is

$φ_{curved} = \frac{q}{ε_{0}} \frac{l}{\sqrt{l^{2} + 4 R^{2}}}$

(iii) Net electric flux leaving a closed surface inside a uniform electric field is zero.

$φ_{net} = 0$

(iv) Electric flux through a cube when the charge is placed at its centre:

(a) Net flux through the cube: $φ = \frac{q}{ε_{0}}$

(b) Flux through each face: $φ_{face} = \frac{q}{6 ε_{0}}$

(v) If the charge is placed at the corner of a cube:

(a) Net flux through the cube: $φ = \frac{q}{8 ε_{0}}$

(b) Flux through each face in contact with the charge is zero.

(vi) When the charge is placed at the centre of one of the faces, then flux through the cube is $φ = \frac{q}{2 ε_{0}}$ .

(vii) Electric flux leaving half-cylindrical surface in a uniform electric field:

$φ = E \times 2 R H$

(viii) Electric flux leaving the conical surface in a uniform electric field:

$φ = E R h$

(ix) Electric flux through a hemisphere in a uniform electric field:

(a) Flux entering the flat face: $φ_{flat} = - π R^{2} E$

(b) Flux leaving the curved face: $φ_{curved} = + π R^{2} E$

(x) Electric flux through an open cylinder:

(a) When a point charge $q$ is placed at the centre of the open side of the cylinder, the flux passing through the cylinder is $φ = \frac{q}{2 ε_{0}}$ .

(b) When a point charge $q$ is placed just above the open side of the cylinder, the flux passing through the cylinder is $φ < \frac{q}{2 ε_{0}}$ .

(c) When a point charge $q$ is placed just inside the open side of the cylinder, the flux passing through the cylinder is $φ > \frac{q}{2 ε_{0}}$ .

24. Electric Potential:

The amount of work done to move a unit positive charge from infinity to a certain point against the electric field.

(i) $V = \frac{W_{ext}}{q_{0}}$

(ii) Electric potential is a scalar.

(iii) S.I. unit: $J C^{- 1}$ .

(iv) Dimension: ${ML}^{2} T^{- 3} A^{- 1}$ .

(v) Positive charge gives positive potential and negative charge produces a negative potential.

25. Electric Potential Due to a Point Charge:

Electric potential due to a point charge $q$ at a distance $r$ is $V = \frac{1}{4 π ε_{0}} \frac{q}{r}$ .

26. Electric Potential Difference:

The amount of work done to move a unit positive charge from one point to another against the electric field.

$Δ V = \frac{W_{ext}}{q_{0}} (or) V_{B} - V_{A} = \frac{W_{A \to B ()}}{q_{0}}$

27. Consider two points $A$ and $B$ at distances $r_{1}$ and $r_{2} (r_{1} < r_{2})$ from a point charge $q$ , then

(i) $V_{A} - V_{B} = \frac{q}{4 π ε_{0}} (\frac{1}{r_{1}} - \frac{1}{r_{2}})$

(ii) Potential decreases, if we move away from positive charge, i.e., $V_{B} - V_{A}$ is $- ve$ , if $q$ is $+ ve$ .

(iii) Potential increases, if we move away from a negative charge, i.e., $V_{B} - V_{A}$ is $+ ve$ , if $q$ is $- ve$ .

(iv) Work done to move a charge $q_{0}$ from $B$ to $A$ is,

$W_{ext} = q_{0} (V_{A} - V_{B}) = \frac{q q_{0}}{4 π ε_{0}} (\frac{1}{r_{1}} - \frac{1}{r_{2}})$

(v) $W_{electric} = - W_{ext}$

28. Electric Potential Due to a System of Charges:

The potential at a point is equal to the scalar sum of potential due to individual charges.

Potential at $P$ is $V = \frac{1}{4 π ε_{0}} (\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}} + \dots + \frac{q_{n}}{r_{n}})$

29. Potential Due to Infinite Charges on $x$ -Axis:

(i) When identical like charges are placed as shown, then the potential at origin is

$V = \frac{1}{4 π ε_{0}} [1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots] = \frac{q}{2 π ε_{0}}$

(ii) If the alternative charges are opposite, then

$V = \frac{q}{6 π ε_{0}}$

30. Potential Due to a Uniformly Charged Ring:

(i) At the centre of the ring, $V_{centre} = \frac{q}{4 π ε_{0} R} = \frac{λ}{2 ε_{0}}$ , where $q$ is charge on the ring, $R$ is radius and $λ$ is linear charge density.

(ii) Potential at a point on the axis:

$V_{axis} = \frac{Q}{4 π ε_{0} \sqrt{R^{2} + x^{2}}} = \frac{λ R}{2 ε_{0} \sqrt{R^{2} + x^{2}}}$

31. Electric Potential Due to a Charged Rod:

(i) On axis:

$V = \frac{Q}{4 π ε_{0} l} l n (1 + \frac{l}{r}) = \frac{λ}{4 π ε_{0}} l n (\frac{r + l}{r})$

(ii) On perpendicular bisector:

$V = \frac{Q}{4 π ε_{0} l} l n [\frac{(\frac{l}{2}) + \sqrt{{(\frac{l}{2})}^{2} + x^{2}}}{(- \frac{l}{2}) + \sqrt{{(\frac{l}{2})}^{2} + x^{2}}}]$

32. Electric Potential Due to a Uniformly Charged Disc on Its Axis:

(i) $V = \frac{σ}{2 ε_{0}} [\sqrt{R^{2} + x^{2}} - x]$

(ii) If $R ≫ x,$ then $V = \frac{σ R}{2 ε_{0}}$ considering potential at $\infty$ as zero.

33. Potential Due to Infinite Line Charge:

(i) $V = \frac{W_{ext}}{q_{0}} = - \int_{\infty}^{r} \vec{E} \cdot d \vec{r} = - \int_{\infty}^{r} \frac{λ}{2 π ε_{0} r} d r$ (not defined) potential at a point due to infinite line charge cannot be determined, but potential difference can be calculated.

(ii) $Δ V = V_{A} - V_{B} = \frac{λ}{2 π ε_{0}} l n (\frac{r_{2}}{r_{1}})$

34. Electric Potential on the Edge of a Uniformly Charged Disc:

$V = \frac{σ R}{2 ε_{0}}$

35. Electric Potential at Centre of a Uniformly Charged Hemispherical Shell:

$V_{center} = \frac{σ R}{2 ε_{0}}$

36. Electric Potential Due to a Uniformly Charged Non-conducting Sphere:

(i) Potential outside the sphere,

$V_{out} = \frac{Q}{4 π ε_{0} x} (x > R)$ or $V_{out} = \frac{ρ R^{3}}{3 ε_{0} x}$

(ii) Potential on the surface of sphere,

$V_{sur} = \frac{Q}{4 π ε_{0} R}$ or $V_{sur} = \frac{ρ R^{2}}{3 ε_{0}}$

(iii) Potential inside the sphere is $V_{inside} = \frac{ρ}{3 ε_{0}} [1.5 R^{2} - 0.5 r^{2}]$ , where $R =$ radius distance and $ρ =$ volume charge density.

(iv) Potential at the centre is $V_{centre} = \frac{3}{2} V_{surface}$ .

37. Potential Due to a Uniformly Charged Spherical Shell (or) Conducting Sphere:

(i) Potential inside the shell is equal to potential on the surface,

$V_{inside} = V_{sur} = \frac{1}{4 π ε_{0}} \frac{Q}{R} = \frac{σ R}{ε_{0}}$

(ii) Potential outside the shell,

$V_{out} = \frac{Q}{4 π ε_{0} x} \frac{σ R^{2}}{ε_{0} x} (x > R)$

38. Potential Difference Between Two Concentric Shells:

Consider two shells of radial $r_{1}$ and $r_{2} (> r_{1})$ are concentric and have charges $q_{1}$ and $q_{2}$ . Then,

(i) Potential of shell $A$ is $V_{A} = \frac{q_{1}}{4 π ε_{0} r_{1}} + \frac{q_{2}}{4 π ε_{0} r_{2}}$ .

(ii) Potential of shell $B$ is $V_{B} = \frac{q_{1} + q_{2}}{4 π ε_{0} r_{2}}$ .

(iii) $V_{B} - V_{A} = \frac{q_{1}}{4 π ε_{0}} (\frac{1}{r_{2}} - \frac{1}{r_{1}})$

(a) If $q_{1}$ is $+ ve$ , then $V_{B} - V_{A} i s - v e (or) V_{A} > V_{B}$ .

(b) If $q_{1}$ is $- ve$ , then $V_{A} < V_{B}$ .

(d) If the two shells are connected by a conducting inner to outer shell. This is the principle of Van de Graff generator.

39. Potential Due to Three Concentric Shells:

The figure shows three conducting concentric shell of radii $a, b$ and $c$ $(a < b < c)$ having charges $Q_{a}, Q_{b}$ and $Q_{c}$ , respectively.

(i) Potential at $A$ is $V_{A} = \frac{1}{4 π ε_{0}} [\frac{Q_{a}}{a} + \frac{Q_{b}}{b} + \frac{Q_{c}}{c}]$ .

(ii) Potential at $B$ is $V_{B} = \frac{1}{4 π ε_{0}} [\frac{Q_{a}}{b} + \frac{Q_{b}}{b} + \frac{Q_{c}}{c}]$ .

(iii) Potential at $C$ is $V_{C} = \frac{1}{4 π ε_{0}} [\frac{Q_{a}}{c} + \frac{Q_{b}}{c} + \frac{Q_{c}}{c}]$ .

40. Earthing of Shell:

The figure shows two concentric spheres having radii $r_{1}$ and $r_{2}$ , respectively, $(r_{2} > r_{1})$ . Charge on the inner sphere is $+ Q$ .

(i) When the outer sphere is earthed:

(a) Let $Q'$ be the charge on the outer surface, then

The potential at the surface of outer sphere, $V_{2} = \frac{1}{4 π ε_{0}} \frac{Q}{r_{2}} + \frac{1}{4 π ε_{0}} \frac{Q'}{r_{2}} = 0$

$\Rightarrow Q' = - Q$

(b) Potential of the inner sphere is $V_{1} = \frac{1}{4 π ε_{0}} \frac{Q}{r_{1}} + \frac{1}{4 π ε_{0}} \frac{(- Q)}{r_{2}} = \frac{Q}{4 π ε_{0}} [\frac{1}{r_{1}} - \frac{1}{r_{2}}]$ .

(ii) In the above case, if the outer sphere is given a charge $+ Q$ and the inner sphere is earthed, then

(a) In this case, the potential at the surface of the inner sphere is zero. So, if $Q'$ is the charge induced on the inner sphere,

then $V_{1} = \frac{1}{4 π ε_{0}} [\frac{Q'}{r_{1}} + \frac{Q}{r_{2}}] = 0$ i.e., $Q' = - \frac{r_{1}}{r_{2}} Q$

(Charge on the inner sphere is less than that of the outer sphere)

(b) Potential at the surface of outer sphere:

$V_{2} = \frac{1}{4 π ε_{0}} \frac{Q'}{r_{2}} + \frac{1}{4 π ε_{0}} \frac{Q}{r_{2}}$

$\Rightarrow V_{2} = \frac{1}{4 π ε_{0} r_{2}} [- Q \frac{r_{1}}{r_{2}} + Q]$ $= \frac{Q}{4 π ε_{0} r_{2}} [1 - \frac{r_{1}}{r_{2}}]$

41. Relation Between Electric Field and Potential Difference:

(i) If $d V$ is the potential difference between two points separated by a distance $d r$ in a uniform electric field on a straight line parallel to the field, then

$E = - \frac{d V}{d r}$ .

(ii) In the above relation, negative sign indicates that the potential decreases if we move in the direction of the electric field.

(iii) Negative of the slope of the $V - r$ graph denotes the intensity of electric field, i.e., $tan θ = \frac{d V}{d r} = - E$ .

(iv) If the potential is a function of the coordinates $x, y, and z$ , then the electric field in space can be written as

$\vec{E} = E_{x} \hat{i} + E_{y} \hat{j} + E_{z} \hat{k}$ , where

$E_{x} = - \frac{\partial V}{\partial x},$ $E_{y} = - \frac{\partial V}{\partial y}$ and

$E_{z} = - \frac{\partial V}{\partial z}$ .

(v) If $d V$ is the potential difference between two points separated by $d \vec{r}$ in a uniform electric field on a straight line making some angle to the field, then the electric field can be determined by knowing the boundary conditions,

$d V = - \int_{r_{1}}^{r_{2}} \vec{E} \cdot d \vec{r} = - \int_{r_{1}}^{r_{2}} E d r cos θ$

(vi) If an electric field is given in space, then the potential difference between two points in the space can be calculated as,

$∆ V = V_{B} - V_{A} = - \int_{r_{1}}^{r_{2}} \vec{E} \cdot d \vec{r} = - \int_{x_{1}}^{x} E_{x} d x - \int_{y_{1}}^{y_{2}} E_{y} d y - \int_{z_{1}}^{z_{2}} E_{z} d z$

42. Equipotential Surface:

The locus of all points that are at the same potential is called the equipotential surface.

Regarding equipotential surface, the following points should be kept in mind:

(i) The direction of the electric field is perpendicular to the equipotential surfaces or lines.

(ii) The equipotential surfaces produced by a point charge or a spherically charge distribution are a family of concentric spheres.

(iii) For a uniform electric field, the equipotential surfaces are planes perpendicular to the field lines.

(iv) Equipotential surfaces due to electric dipole and two identical charges separated are shown in the below figure.

(v) A metallic surface of any shape is an equipotential surface.

(vi) Equipotential surfaces can never cross each other.

(vii) The work done in moving a charge along an equipotential surface is always zero.

43. Electric Dipole:

Two equal and opposite charges separated by a small distance.

(i) Electric dipole moment $(\vec{p})$ :

The product of charge and length of the dipole.

(a) Dipole moment, $\vec{p} = q \times 2 \vec{l}$

(b) S.I. unit: $C m$

(d) It is a vector quantity.

(e) Its direction is always from $- q$ to $+ q$ .

(ii) Electric field on axial line:

(a) ${\vec{E}}_{axial} = \frac{1}{4 π ε_{0}} \frac{2 p r}{(r^{2} - l^{2})} \hat{r}$

(b) For a short dipole:

${\vec{E}}_{axial} = \frac{1}{4 π ε_{0}} \frac{2 \vec{p}}{r^{3}}$

(iii) Electric field on equatorial line:

(a) ${\vec{E}}_{equatorial} = \frac{1}{4 π ε_{0}} \frac{(- \vec{P})}{{(r^{2} + l^{2})}^{3 / 2}}$

(b) For short dipole, ${\vec{E}}_{e q u a t o r i a l} = \frac{- \vec{p}}{4 π ε_{0} r^{3}}$ .

(iv) Electric field at the centre of dipole:

${\vec{E}}_{mid} = \frac{1}{4 π ε_{0}} \frac{2 q}{l^{2}} = \frac{1}{4 π ε_{0}} \frac{\vec{p}}{l^{3}}$

(v) Electric field at any general point due to a short dipole:

(a) $E = \frac{p}{4 π ε_{0} r^{3}} \sqrt{1 + 3 c o s^{2} θ}$

(b) If $ϕ$ is the angle made by the electric field with position vector $\vec{r}$ , then $t a n ϕ = \frac{1}{2} tan θ$ .

(vi) Electric potential at midpoint of the dipole, $V_{mid} = 0$ .

(vii) Potential on axial line, $V_{axial} = \frac{k p}{r^{2} - l^{2}}$ .

For $r ≫ l, V_{axial} = \frac{1}{4 π ε_{0}} \frac{P}{r^{2}}$

(viii) Potential at any point on an equatorial plane is zero.

(ix) Potential at any general point due to dipole is,

$V_{g e n e r a l} = \frac{1}{4 π ε_{0}} \frac{p cos θ}{r^{2}}$

44. Dipole in a Uniform Electric Field:

(i) Net force on the dipole is zero.

(ii) Torque on dipole is $\vec{r} = \vec{p} \times \vec{E}$ .

$\Rightarrow |\vec{τ}| = p E s in θ$ , where $θ$ is the angle between electric field and dipole moment.

45. Dipole in a Non-uniform Electric Field:

(i) Force on the dipole, $\vec{F} = - \vec{p} \cdot \frac{d \vec{E}}{d x}$

(ii) Torque on the dipole is $\vec{τ} = \vec{l} \times ({\vec{E}}_{1} + {\vec{E}}_{2})$ , where $E_{1}$ and $E_{2}$ are electric fields at the point charges.

46. Electrostatic Potential Energy:

(i) Potential energy of the two-particle system in an external electric field:

$U = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r}$

(ii) Potential energy of the two-particle system in an external electric field:

$U = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r} + q_{1} V_{A} + q_{2} V_{B}$

(iii) Potential energy of three particle system,

$U = \frac{k q_{1} q_{2}}{r_{12}} + \frac{k q_{2} q_{3}}{r_{23}} + \frac{k q_{3} q_{1}}{r_{31}}$

(iv) If the system has $n$ particles, then the number of terms in potential energy of the system is $\frac{n (n - 1)}{2}$ .

(v) Potential energy of short dipole in a uniform electric field is,

$U = - \vec{p} \cdot \vec{E} = - p E cos θ$

(a) If $θ = 0 °, U = - p E$ (minimum). The dipole is in stable equilibrium.

(b) If $θ = 180 °$ , $U = + p E$ (maximum). The dipole is in unstable equilibrium.

47. Equilibrium of Charged Soap Bubble:

(i) For a charged soap, bubble of radius $R$ and surface tension $T$ and charge density $σ .$ The pressure due to surface tension $\frac{4 T}{R}$ and atmospheric pressure $P_{out}$ act radially inwards and the electrical pressure $(P_{el})$ acts radially outward.

(a) The total pressure inside the soap bubble:

$P_{in} = P_{out} + \frac{4 T}{R} - \frac{σ^{2}}{2 ε_{0}}$

(b) Excess pressure inside the charged soap bubble:

$P_{in} - P_{out} = P_{excess} = \frac{4 T}{R} - \frac{σ^{2}}{2 ε_{0}}$

(ii) If air pressure inside and outside is assumed equal, then

$P_{in} = P_{out}$ , i.e., $P_{excess} = 0$ . So, $\frac{4 T}{R} = \frac{σ^{2}}{2 ε_{0}}$

(a) Charge density: Since $\frac{4 T}{R} = \frac{σ^{2}}{2 ε_{0}}$ ⇒ $σ = \sqrt{\frac{8 ε_{0} T}{R}} = \sqrt{\frac{2 T}{π k R}}$

(b) Radius of bubble: $R = \frac{8 ε_{0} T}{σ^{2}}$

(d) Total charge on the bubble: $Q = 8 π R \sqrt{2 ε_{0} T R}$

(e) Electric field intensity at the surface of the bubble:

$E = \sqrt{\frac{8 T}{ε_{0} R}} = \sqrt{\frac{32 π k T}{R}}$

(f) Electric potential at the surface: $V = \sqrt{32 π R T k} = \sqrt{\frac{8 R T}{ε_{0}}}$

48. Self-Energy:

(i) Self-energy of a hollow sphere, $U = \frac{Q^{2}}{8 π ε_{0} R}$

(ii) Self-energy of a hollow sphere, $U = \frac{{3 Q}^{2}}{20 π ε_{0} R}$

49. Interaction energy of a system is the sum of self-energies of the bodies and electrostatic interaction potential energies between the bodies.

50. Charge Distribution on Parallel Metal Plates:

(i) Charge on the outermost surfaces is equal to half the sum of the charges on all the plates.

(ii) Charges on the opposite plates are equal and opposite.

(iii) Example:

In this figure,

(a) Charges on the outer surfaces $a$ and $f$ are the same and equal to $\frac{q_{1} + q_{2} + q_{3}}{2}$ .

(b) Charges $q_{b} = - q_{c}$ and $q_{d} = - q_{e}$ .

The electric field in a region is given as, E→=10i^+20j^ V m-1 The net flux passing through a square area of side 2 m, parallel to x-z, plane is,

Important Questions on Electrostatics

Learn and Prepare for any exam you want !

The electric field in a region is given as, $\vec{E} = (10 \hat{i} + 20 \hat{j}) V m^{- 1}$ The net flux passing through a square area of side $2 m$ , parallel to $x - z$ , plane is,