physics_electrostatics
 Electrostatics Electrostatic Charge It is a deficiency or excess of electrons which happens on insulating surfaces. Negative ChargeExcess of electron Positive ChargeDeficiency of electron Basic Properties of electric charge1. Additivity of charges2. Charge is conserved3. Quantisation of charge Unit of charge: Coulomb-1C = 6×1018$6×{10}^{18}$ electrons (approx.) When we rub a glass rod with silk, some of the electrons from the rod are transferred to the silk cloth. Thus the rod gets positively charged and the silk gets negatively charged. ConductorsThose which allow electricity to pass through them easily are called conductors. They have electric charges (electrons) that are free to move inside the material. InsulatorsThose which offer high resistance to the passage of electricity through them called insulators. When some charge is transferred to a conductor, it readily gets distributed over the entire surface of the conductor. In contrast, if some charge is put on an insulator, it stays at the same place. The interior of a conductor can have no excess charge in the static situation Grounding or Earthing The process of sharing the charges with the earth is called grounding or earthing. Coulomb's Law:Force between two point charges varies inversely as the square of the distance between the charges and directly proportional to the product of the magnitude of the two charges andacted along the line joining the two charges. F=14𝜋𝜖0 q1q2r2$F=\frac{1}{4𝜋{𝜖}_{0}}\phantom{\rule{0.22em}{0ex}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$ Forces due to multiple charge Electric Field Linesa. Electric field lines are a way of pictorially mapping the electric field around a configuration of charges.b. Electric field lines are curve in three dimensions drawn in such a way that the tangent to it at each point is in the direction of the net field at that point. c. An arrow on the electric field lines specify the direction of electric field.d. Relative density (i.e., closeness) of the field lines at different points indicates the relative strength of electric field at those points.e. Electrostatic field lines do not form any closed loops.f. Electric Field lines start from positive charges and end at negative charges. If there is a single charge, they may start or end at infinity.g. In a charge-free region, electric field lines can be taken to be continuous curves without any breaks.h. Two field lines can never cross each other. (If they did, the field at the point of intersection will not have a unique direction, which is absurd.)i. At the surface of a charged conductor, electrostatic field must be normal to the surface at every point. Electric fieldElectric field due to a charge Q at a point in space may be defined as the force that a unit positive charge would experience if placed at that point. Electric field due to point charge Q at a distance r from QE(r) = 14𝜋𝜖0Qr2$E\left(r\right)\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}\frac{1}{4𝜋{𝜖}_{0}}\frac{Q}{{r}^{2}}$Unit of electric field=NC$=\frac{N}{C}$ Electric field due to system of charges Physical significance of electric fielda. Electric field is an elegant way of characterising the electrical environment of a system of charges. b. Electric field at a point in the space around a system of charges tells you the force a unit positivetest charge would experience if placed at that point (without disturbing the system). c. Electric field is a characteristic of the system of charges and is independent of the test charge that you place at a point to determine the field. d. Electric field is a vector field, since force is a vector quantity. 1. Faraday developed idea of electric field lines of force.2. Electric lines of force from small surface area of a positively charged conductor enclose a tube like structure and is called tube of force.3. Tube of forces possesses same properties as exhibited by electric lines of force.4. Only one tube of induction startes from unit positive charge.5. Number of tubes of induction from positive charge q = q6. Number of tubes of induction passing normally per unit area at distance r from positive charge q =q4𝜋r2=𝜖(q4𝜋𝜖r2)=𝜖E$=\frac{q}{4𝜋{r}^{2}}=𝜖\left(\frac{q}{4𝜋𝜖{r}^{2}}\right)=𝜖E$ Electric Flux Electric flux measures the electric field through a given surface (surface area).𝛷E=𝛴 E.𝛥s${𝛷}_{E}=𝛴\phantom{\rule{0.22em}{0ex}}\stackrel{\to }{E}.\stackrel{\to }{𝛥s}$ Electric dipole : An electric dipole is a pair of equal and opposite point charges q and –q, separated by a distance 2a. The line connecting the two charges defines a direction in space. By convention, the direction from –q to q is said to be the direction of the dipole. The mid-point of locations of –q and q is called the centre of the dipole. Dipole moment of an electric dipole =p=2qa$=p=2qa$ Electric field due to dipole on a point at axisE= q4𝜋𝜖04ar(r2-a2)2$E=\phantom{\rule{0.22em}{0ex}}\frac{q}{4𝜋{𝜖}_{0}}\frac{4ar}{\left({r}^{2}-{a}^{2}{\right)}^{2}}$where r is distance of point from center of dipole.E≈ 4 q a4𝜋𝜖0r3 if r≫a$E\approx \phantom{\rule{0.22em}{0ex}}\frac{4\phantom{\rule{0.22em}{0ex}}q\phantom{\rule{0.22em}{0ex}}a}{4𝜋{𝜖}_{0}{r}^{3}}\phantom{\rule{1.32em}{0ex}}if\phantom{\rule{0.22em}{0ex}}r\gg a$ Electric field due to dipole on a point at equitorial plane aE= 2qa4𝜋𝜖0 (r2+a2)3/2 E≈ 2 q a4𝜋𝜖0r3 if r≫a $\begin{array}{l}E=\phantom{\rule{0.22em}{0ex}}\frac{2qa}{4𝜋{𝜖}_{0}\phantom{\rule{0.22em}{0ex}}\left({r}^{2}+{a}^{2}{\right)}^{3/2}}\\ \\ E\approx \phantom{\rule{0.22em}{0ex}}\frac{2\phantom{\rule{0.22em}{0ex}}q\phantom{\rule{0.22em}{0ex}}a}{4𝜋{𝜖}_{0}{r}^{3}}\phantom{\rule{1.32em}{0ex}}if\phantom{\rule{0.22em}{0ex}}r\gg a\\ \end{array}$ Dipole in a uniform electric field Net force F = 0 (zero) Torque𝜏=p×E $\stackrel{\to }{𝜏}=\stackrel{\to }{p}×\stackrel{\to }{E}\phantom{\rule{0.22em}{0ex}}$ Continuouus charge distribution1. Linear charge density2. Surface charge density3. Volume charge density Electric Flux Electric flux measures the electric field through a given surface (surface area).𝛷E=𝛴 E.𝛥s${𝛷}_{E}=𝛴\phantom{\rule{0.22em}{0ex}}\stackrel{\to }{E}.\stackrel{\to }{𝛥s}$ Gauss’s lawElectric flux through a closed surface S𝛷E=q𝜖0 ${𝛷}_{E}=\frac{q}{{𝜖}_{0}}\phantom{\rule{0.44em}{0ex}}$ where q = total charge enclosed by S. Electric field due to infinitely long straight uniformly charged wire at distance rE=𝜆2𝜋𝜖0r$E=\frac{𝜆}{2𝜋{𝜖}_{0}r}$ Electric field due to infinite plane sheet at distance dE=𝜎2𝜖0$E=\frac{𝜎}{2{𝜖}_{0}}$ Electric field inside the shell is zero Electric field at the surface of a charged conductor=E=𝜎𝜖0$=E=\frac{𝜎}{{𝜖}_{0}}$ Polarisation : The dipole moment per unit volume is called polarisation. P=𝜒eE $P={𝜒}_{e}E\phantom{\rule{0.22em}{0ex}}$𝜒e${𝜒}_{e}$=electric susceptibility of the dielectric medium.𝜒e${𝜒}_{e}$ is molecular property of the substance Electrostaic potential is constant throught out the volume of the conductor and has the same value (as inside) on its surface Potential due to electric dipoleV= q4𝜋𝜖02 a cos𝜃r2=p cos𝜃4𝜋𝜖0 r2$V=\phantom{\rule{0.22em}{0ex}}\frac{q}{4𝜋{𝜖}_{0}}\frac{2\phantom{\rule{0.22em}{0ex}}a\phantom{\rule{0.22em}{0ex}}cos𝜃}{{r}^{2}}=\frac{p\phantom{\rule{0.22em}{0ex}}cos𝜃}{4𝜋{𝜖}_{0}\phantom{\rule{0.22em}{0ex}}{r}^{2}}$ Inside a conductor electrostatic field is zero Total Normal Electric Induction (TNEI) over any closed surface is equal to algebraic sum of charges enclosed by that surface. Potential due to point charge Q at a distance r from Q= V(r) = 14𝜋𝜖0Qr $=\phantom{\rule{0.22em}{0ex}}V\left(r\right)\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}\frac{1}{4𝜋{𝜖}_{0}}\frac{Q}{r}\phantom{\rule{0.22em}{0ex}}$ An equipotential surface is a surface with a constant value of potential at all points on the surface.passing through that point. For any charge configuration, equipotential surface through a point is normal to the electric field at that point.|E|=-dVdr$|E|=-\frac{dV}{dr}$ (i) Electric field is in the direction in which the potential decreases steepest.(ii) Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point. Equipotential surface An equipotential surface is a surface with a constant value of potential at all points on the surface. Electric field at the surface of charged conductorE=𝜎𝜖0$E=\frac{𝜎}{{𝜖}_{0}}$ where 𝜎=$𝜎=$Surface charge density Mechanical force acting on unit area of a charged conductor (f). It is also called electrostatic pressure.f=𝜎22𝜖=𝜎22k𝜖0 $f=\frac{{𝜎}^{2}}{2𝜖}=\frac{{𝜎}^{2}}{2k{𝜖}_{0}}\phantom{\rule{0.22em}{0ex}}$ The capacitance C of the parallel place capacitor=C=QV=𝜖0Ad$=C=\frac{Q}{V}=\frac{{𝜖}_{0}A}{d}$Unit of capacitance = Farad Energy stored in a capacitor=U=12CV2=12QV=Q22C$=U=\frac{1}{2}C{V}^{2}=\frac{1}{2}QV=\frac{{Q}^{2}}{2C}$ Energy density of electric field per unit volume=u=12𝜖0E2 $=u=\frac{1}{2}{𝜖}_{0}{E}^{2}\phantom{\rule{0.22em}{0ex}}$ Dielectric constant=KK=𝜖𝜖0$K=\frac{𝜖}{{𝜖}_{0}}$where𝜖0=${𝜖}_{0}=$permittivity of vaccum 𝜖=$𝜖=$permittivity of medium Dielectric constant=KK=CC0$K=\frac{C}{{C}_{0}}$ Capacitance in parallel (Equivalent capacitance=C)C=C1+C2+C3$C={C}_{1}+{C}_{2}+{C}_{3}$ Capacitance in series(Equivalent capacitance=C)1C$\frac{1}{C}$=1C1$\frac{1}{{C}_{1}}$+1C2$\frac{1}{{C}_{2}}$+1C3$\frac{1}{{C}_{3}}$
 QAa. A dipole placed in a uniform electric field experiences only a torqueb. Once the dipole is aligned in the direction of field, torque vanishes QAa. In non-uniform field, electric dipole expereinces both torque and force. QState Coulomb’s law in electrostatics.ACoulomb's Law:Force between two point charges varies inversely as the square of the distance between the charges and directly proportional to the product of the magnitude of the two charges andacted along the line joining the two charges. F=14𝜋𝜖0 q1q2r2$F=\frac{1}{4𝜋{𝜖}_{0}}\phantom{\rule{0.22em}{0ex}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$ Q.Can two equipotential surface intersect each other ? Justify.A.On an equipotential surface, all the points have same potential. So we can say potential difference between any two points is zero on equipotential surface.Electric field is always perpendicular to the equipotential surface. Now if two equipotential surface intersects then there will be two electric fields perpendicular to their respective equipotential surface. But there cannot be two directions of the electric field at any single point (intersection point) So two equipotential surface cannot intersect each other.
 Questions Q.1.19A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?A.1.19Gauss’s lawElectric flux through a closed surface S𝛷E=q𝜖0 ${𝛷}_{E}=\frac{q}{{𝜖}_{0}}\phantom{\rule{0.44em}{0ex}}$where q = total charge enclosed by S. 𝛷E=2×10-68.854×10-12=2.258×105 Nm2C-1 ${𝛷}_{E}=\frac{2×{10}^{-6}}{8.854×{10}^{-12}}=2.258×{10}^{5}\phantom{\rule{0.22em}{0ex}}N{m}^{2}{C}^{-1}\phantom{\rule{0.22em}{0ex}}$ Q.1.22A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?A.1.22(a)Charge on sphereQ=(4𝜋r2)×80×10-6$Q=\left(4𝜋{r}^{2}\right)×80×{10}^{-6}$Q=(4 𝜋 1.22)×80×10-6$Q=\left(4\phantom{\rule{0.22em}{0ex}}𝜋\phantom{\rule{0.22em}{0ex}}{1.2}^{2}\right)×80×{10}^{-6}$Q=1.447×10-3 C $Q=1.447×{10}^{-3}\phantom{\rule{0.22em}{0ex}}C\phantom{\rule{0.22em}{0ex}}$ (b)total electric flux leaving the surface of the sphere𝛷E=q𝜖0=1.447×10-38.854×10-12 ${𝛷}_{E}=\frac{q}{{𝜖}_{0}}=\frac{1.447×{10}^{-3}}{8.854×{10}^{-12}}\phantom{\rule{0.22em}{0ex}}$ ⟹𝛷E=q𝜖0=1.447×10-38.854×10-12 $⟹{𝛷}_{E}=\frac{q}{{𝜖}_{0}}=\frac{1.447×{10}^{-3}}{8.854×{10}^{-12}}\phantom{\rule{0.22em}{0ex}}$ ⟹𝛷E=1.63×108 Nm2/C $⟹{𝛷}_{E}=1.63×{10}^{8}\phantom{\rule{0.22em}{0ex}}N{m}^{2}/C\phantom{\rule{0.22em}{0ex}}$