12th Electrostatics PDF

# ELECTROSTATICS ## Introduction:- Electrostatics deals with the study of electric charges at rest. Electric charge produces electric field. Electric field is a characteristic of system of charges. ## Electric lines of force:- Michael Faraday developed idea of electric lines of force. Electric...

# ELECTROSTATICS ## Introduction:- Electrostatics deals with the study of electric charges at rest. Electric charge produces electric field. Electric field is a characteristic of system of charges. ## Electric lines of force:- Michael Faraday developed idea of electric lines of force. Electric lines of force is defined as the path along which (free) unit positive charge moves. ## Electric flux - It is the product of area and the component of electric intensity normal to the surface. The number of tubes of forces passing normally through the given surface is called electric flux through that area. SI Unit:- Volt-meter ## Electric field Intensity:- The electric flux per unit area around a point is called electric intensity at that point. ie. electric flux density. SI Unit: v/m or N/C ## Gauss' Law:- Gauss' law states that the total electric flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ε₀. Φ = ∫ E . ds = q/ε₀ Where, q: total charge Φ: total electric flux ## Application's of Gauss' Law:- ### 1) Electric field Intensity due to Uniformly charged spherical shell or Hollow sphere:- Consider a sphere of radius R carrying a positive charge +q placed in a dielectric medium of permitivity ε (ε = ε₀k). Let P be a point at distance r (r>R) from centre of charged sphere. Uniformly charged spherical shell or hollow sphere To determine the electric field intensity at P, imagine a concentric Gaussian sphere of radius r passing through P, as shown in fig. Let ds be a small area around point P on Gaussian surface. Due to symmetry and spheres being concentric, electric field at each point on the Gaussian surface has same magnitude E & directed radially outwards. Angle beth direction of E and normal to surface of sphere (ds) is zero. .. cosθ = 1 ∴ E . ds = Eds cosθ = Eds flux dΦ through ds = Eds But total electric flux through Gaussian surface is Φ Φ = ∫ E . ds = ∫ Eds = E ∫ds = E 4πr² [ ∫ds = Area of sphere= 4πr²] But, According to Gauss theorem Φ = q/ε₀ = ∫ E . ds - (2) From eqn (1) & (2) q/ε₀ = E 4πr² E = q/4πε₀r² - (3) If σ (sigma) is the surface density of charge of sphere then, σ = q/Area q = σ × 4πR² Put eqn ④ in eqn ③, we get, E = 4πR²/4πε₀r² E = σR²/ε₀r² - (4) From eqn (3) it is seen that, the electric field at a point outside the shell is same as that due to point charge. Hence, it can be concluded that a uniformly charged sphere is equivant to a point charge at its centre. ### Case I:- If point P lies on the surface of charged sphere ∴ r=R E = q/4πε₀R² = σ/ε₀ ### Case II:- If point P lies inside the sphere, then σ=0 because inside sphere there are no charges. E=0 ### 2) Electric field Intensity due to an Infinitely Long straight charged Wire (Cylinder):- Consider a uniformly charged wire A of infinite length having a constant linear charge density (charge per unit length), λ, kept in medium of permittivity ε (ε=ε₀k). Let P be a point at distance r from axis of cylinder. To determine intensity at point P, imagine a coaxial Gaussian cylinder B of length l and radius r passing through point P. Let ds bet the small area at point P on Gaussian surface. Let E be electric field intensity at P. By symmetry, E is directed outwards at right angles to axis of cylinder. & its magnitude is same at every point on Gaussian cylinder. Here, angle between direction of E and normal to ds is zero. .. cos θ = cos 0° = 1 .. E . ds = Eds cos θ = Eds .. Flux dΦ through ds = Eds But total electric flux through Gaussian surface Φ is given by, Φ = ∫ E . ds = ∫ Eds = E ∫ds Φ = E 2πrl - (1) [∫ds = Area = 2πrl ] But From Gauss law, Φ = q/ε₀ - (2) .. From egn① & ②, we get, q/ε₀ = E 2πrl q/ε₀ = E 2πrd - (3) But λ = q/l λ = q ⇒ q = λl Put q=λl in eqn ③ λl = E 2πrd E = λ/2πε₀r # Electrostatic potential energy- Electrostatic potential energy is the work done against the electrostatic forces to achieve a certain configuration of charges in a given system. For bringing like charges (ie. two positive) close to each other, work is necessary. This work done is equal to change in potential energy of their system. : Work done against electrostatic force = dU : F . dr = dU Where, dU = increase in potential energy dr= displacement of charge F = force exerted on charge. ## Expression for potential energy- *Charge +q₀ displaced by dr towards charge +q* Consider the electrostatic field due to a source charge +q placed at origin O. Another small charge +q₀ is taken from point A to point B. Let the distance OA = r₁, & OB = r₂ While displacing the charge q₀ through displacement dr work is done against electrostatic force F_E. It causes increase in potential energy of system. : dU = F_E.dr = -F_E'dr Negative sign indicates that dr & electrostatic force F_E are opposite in direction. Total potential energy as the charge q₀ moves from A to B is given by, ΔU = ∫ dU = ∫-F_E'dr The electrostatic force between two charges seperated by distance r is given by, F_E = [1/4πε₀] (q₀.q/r²) - 3 r is unit vector in direction of r. eqn ③ becomes, ΔU = ∫ dU = ∫[1/4πε₀] (q₀.q)/r²'dr ΔU=- [1/4πε₀] q₀q (1/r₂ - 1/r₁) # Electrostatic potential:- We know that, U(r)=(1/4πε₀) (q₁q₂/r) = (1/4πε₀) (q₂/r²) = (1/4πrε₀) (q₁/r) , In general, V(r)= (q/4πε₀r) depends on the charge q and location of point at distance r from it. This is called electrostatic potential of charge q at distance r from it. Hence in terms of potential, potential energy can be written as, U(r) = V₁ (r) q₂ = V₂ (r) q₁ Where, V₁(r) = Potential of charge q₁ at dist r V₂(r) = " " q₂ " " .. Electrostatic potential (U) = electric potential (V) x charge(q). energy Electrostatic potential (V) = Electrostatic potential energy per unit charge. V = U/q Electrostatic potential difference between any two points in an electric field is V₂-V₁ = (U₂-U₁)/q = dW/q Workdone per unit charge to move charge q from point 2 to point 1. ## Relation between Electric field and electric potential- Consider a point charge +q at point O'. The points M & N are seperated by small distance dч. Let q₀ be a test charge to be placed at point M. The work done to move q₀ from M to N is given by, dW = -F dч Negative sign indicates that force F & displacement are in opposite direction. We know that, Electric intensity E = F/q₀ = q/4πε₀r² .. F = E q₀ .. dW = -E q₀ dч = dV .. F = -dV/dч Hence, the electric field at a point in an electric field is the negative of the potential gradient of that point. ## Zero Potential- The nature of potential is such that its zero point is arbitrary. Once the zero point is set, then every potential is measured with respect to that reference. The zero point is set conveniently. In case of point charge, the zero point is set at infinity, for electrical circuits, the earth is taken to be at zero potential. ## Electric potential due to a point charge- Consider a point charge +q is located at point O'. We have to find potential at point A at distance r from point O. The unit positive charge is brought from infinity to point A. The electric potential at point A is the work done per unit positive charge which is displaced from ∞ to A. This work done is independent of path hence the path is OA extending to ∞ Electric potential due to a point charge Let OM=r₁. The electrostatic force on a unit positive charge at M is F = q/4πε₀ r₁² - ① Now, unit positive charge is displaced from M to N through distance dч against force. Hence work done is given by, dW = -F dч .. Total workdone in displacing unit positive charge from ∞ to A is given by, W = ∫-F dч = ∫-q/4πε₀ r₁² dч -q/4πε₀ ∫r₁^-2dч = -q/4πε₀ [r₁^-1/ -1]^∞_r : [∞ -> 0 ] = -q/4πε₀ [1/∞ - 1/r] W= q/4πε₀r - ③ But, by defn V = q/4πε₀r - ④ It is electrostatic potential at A due to charge q. .. V=W= q/4πε₀r A positively charged particle produces a positive electric potential & negatively charged particle produces negative electric potential. At r=∞ V= q/4πε₀∞ = 0 This shows that electric potential is zero at infinity. The value of V for any point at distance r from point charge q is same, & it is independent of direction of r. Hence, electrostatic potential due to a single charge is spherically symmetric. We know E = q/4πε₀r² Electric intensity E α 1/r² # Electric potential due to an electric dipole:- A pair of equal & opposite charges seperated by a finite distance is called an electric dipole. The line joining centres of two charges is called dipole axis. A straight line drawn perpendicular to the axis and passing through centre O of the electric dipole is called equator of dipole. Consider an electric dipole AB having charge +q at point A & -q at point B seperated by a finite distance 2l. Let O be the centre of dipole. ; OA = OB=l The dipole moment P having magnitude P = 2q×l is directed from -q to +q. Let P be any point near electric dipole at distance op=r from centre O. Join AP & BP. AP= r₁ & BP= r₂. Electric potential at P due to +q, is V₁ = +q/Απε₀r₁ - (1) Electric potential at P due to -q is V₂ = -q/Απε₀r₂ - (2) Net potential at point P due to dipole is V=V₁+V₂ = q/4πε₀r₁ - q/4πε₀r₂ = q/4πε₀(1/r₁ - 1/r₂) - (3) From figure in ΔAOC using cosine rule r₁² = r² + l² - 2rl cosθ r₁² = r²(1+l²/r² - 2l/r cosθ) - (4 ) In Δ BOC; r₂² = r² + l² - 2rl cos(π-θ) r₂² = r² + l² + 2rl cosθ r₂² = r²(1+l²/r² + 2l/r cosθ) - (5) For a short dipole, 2l<<<r If r>>l, l²/r² is small . Neglecting l²/r² eqn (4) & (5) becomes, r₁² = r²(1- 2l/r cosθ) r₂² = r² (1- 2l/r cosθ) .. r₂ = r(1- 2l/r cosθ)^1/2 .. r₂ = r ( 1 + 2l/r cosθ )^1/2 By putting r₁ & r₂ in eqn (3) we get. V_c = q/4πε₀r [ (1 - 2l/r cosθ )^1/2 - ( 1 + 2l/r cosθ )^1/2] = q/4πε₀r [ (1 - 2l/r cosθ )^1/2 - ( 1 + 2l/r cosθ )^1/2] = - (q/4πε₀r) [(1 - 2l/r cosθ )^1/2 - ( 1 + 2l/r cosθ )^1/2 ] - (6) By using Binomial theorem, we know (1+x)ⁿ = (1 + nx), if x<<1 and retaining terms upto first order of 1/r only. (1 + 2l/r cosθ)-1/2 = (1 + (1/2) (-2l/r cosθ)) = (1 - l/r cosθ) ∴ (1- 2l/r cosθ)-1/2 = (1 + l/r cosθ ) (1 + 2l/r cosθ)-1/2 = (1 - l/r cosθ ) .. eqn (6) becomes V_c = q/4πε₀r [ (1 + l/r cosθ) - (1 - l/r cosθ ) ] = q/4πε₀r (2l/r cosθ) V_c = (q2l cosθ)/4πε₀r² - (7) ..[P= q×2l] Electric potential at C, can also be expressed as V_c = 1/4πε₀ (P.r)/r³ V_c = 1/4πε₀ (P.r^{^})/r² ..( r^{^} = r/r) Where, r^{^} is a unit vector along the position vector, OC=r^{^} ### Case 1:- If point lies on axis, θ= 0° or 180° .. eqn ⑦ becomes, V_axial = ±1/4πε₀ * (P/r²) This is maximum value of potential (towards +q). ### If point lies on equator, θ= 90° cos θ = cos 90° = 0 .. V_equatorial = 0 The potential at any point on equatorial line of dipole is zero. This is minimum value of magnitude of potential of dipole. The plane perpendicular to line between charges at the midpoint is equipotential plane having zero potential. The workdone to move a charge anywhere in this plane will be zero. # Electrostatics potential due to system of charges- Consider a system of n charges q₁/q₂/q₃.......q_n at distances r_1, r₂, r_n respectively from point P. The potential V₁ at P due to charge q₁ is V₁ = 1/4πε₀ * (q₁/r₁) Similarly, potentials V₂, V₃.... V_n at P due to charges q₂, q₃, ....... q_n are given by, V₂ = 1/4πε₀ * (q₂/r₂) , V₃ = 1/4πε₀ * (q₃/r₃) & V_n=1/4πε₀ * (q_n/r_n) By superposition theorem V = V₁ + V₂ + ... + V_n = 1/4πε₀ (q₁/r₁ + 1/4πε₀ (q₂/r₂ + .... + 1/4πε₀ (q_n/r_n = 1/4πε₀ ( q₁ /r₁+ q₂ /r₂+ .... + q_n /r_n) V = 1/4πε₀ ∑_i=1ⁿ q_i/r_i ## Equipotential Surface :- An equipotential surface is that surface at every point of which the electric potential is same. The potential (V) for a single charge q is given by, V = 1/4πε₀ * (q/r) If r is constant then V will be constant. Hence equipotential surfaces of single point charge are concentric spherical surfaces centred at the charge. **By def, potential difference beth two points P & q is work done** Vp- V_q = W_pq If P & q lies on equipotential surface, Vp = Vq W_pq =0 Hence, no work is required to move test charge along an equipotential surface. a) If dч is small distance over equipotential surface due to charge. dw = E . dч = Edu cos θ But dw = 0 0 = Edu cos θ .. cos θ = 0 or θ = 90° .. [E⊥ dч] ... electric intensity E is always normal to equipotential surface. b) If the field is not normal, it would have non-zero component along the surface. There is work have to be done to move a test charge against this component. There is no P.D. between any two points on equipotential surface hence work is not required to displace charge on surface. Hence, Electric field must be normal to equipotential surface at every point & vice versa. # Dielectrics & electric Polarisation:- Dielectrics are insulators which can be used to store electrical energy. This is because when such substances are placed in an external field their positive & -ve charges get displaced in opposite directions and the molecules develop a net dipole moment. This is called polarization of the material & such materials are called dielectrics. e.g. Glass, wax, mica, rubber etc. In every atom there is a positively changed nucleus & negatively charged electrons surrounding it." The centre of negative charge is the centre of mass of electrons & centre of positive charge is centre of mass of protons. Dielectrics can be classified as polar dielectrics & non polar dielectrics as described below: - ## a) Polar Dielectric- 1) The molecules in which centre of gravity of +ve nuclei & revolving electrons do not coincide are known as polar molecules. e.g. HCl, H₂O, N₂O, etc. 2) Polar substances behaves as a tiny electric dipole. 3) Polar molecules have a permanent electric dipole moment of order of 10^-30 cm. 4) Water molecule has bent shape with its two O-H bonds which are inclined at an angle of about 105°. 5) It has high dipole moment of 6.1×10^-30 cm. 6) they have asymmetric shapes. ## Non-polar Dielectrics:- 1) The molecules in which centre of gravity of positive nuclei & revolving electrons coincide are known as non - polar molecules. e.g. H₂, O₂, CO₂, polythene etc. 2) Non polar molecules do not have permanent electric dipole moment because of their symmetry. 3) They have symmetrical shapes. # Polarization of a polar dielectric in an external electric field:- In polar dielectric, in absence of electric field, electric dipole, are randomly oriented. Hence, total dipole moments of polar dielectric is zero as shown in fig.a ε₀=0 When substance is subjected to an external electric field, the dipole moments of different, molecules tends to align in the direction of field. As a result the dielectric develops a net dipole moment in the direction of the external field, Hence the dielectric is polarized as shown in fig. ε₀≠0 Polarization of a non-polar dielectric in an external electric field:- In absence of electric field (E=0) the structure of non-polar dielectric is as shown in fig.c) # Polarization:- The dipole moment per unit volume is called polarization. It is denoted by P Polarization = Dipole moment/Volume For linear isotropic dielectrics, P = χ_eE Where, χ_e is a constant called dielectric susceptibility of the dielectric medium. For Vacuum χ_e=0 ## Reduction of electric field due to polarization of a dielectric: When a dielectric is placed in an external electric field the value of field inside the dielectric is less than the external field as a result of polarization. Consider a thin slab of dielectric is placed in uniform external field as shown in figure. Since electric charges are not free to move about in a dielectric, no current passe flows. Instead of moving charges, electric field produce slight changes within atoms, resulting in aligning them with the field. During this, charges moves over distance that are less than atomic diameter. Due to this alignment, there is an apparent sheet of +ve charges on the right side &-ve charges on left side of dielectric. These two sheets of induced surface charges produce an electric field which called polarization field in the insulators which opposes the applied field E. The net field E' inside the dielectric is E' = E-E₀ - (1) # Capacitors & Capacitance ## ⅰ) Capacitor or condenser:- It is an electrical device used for storing electric charges. It consist of two conductors separated by air or dielectric material. ## • Principle:- The presence of an electrical earthed conductor near a charged body, reduces the potential of the charged body & increases its capacity. ## Types of condensers - 1) Parallel plates condenser 2) Spherical condenser 3) Cylindrical condenser ## Capacitance - Capacity (or capacitance) of condenser is defined as its ability to hold the electrical charges. It is observed that the potential of conductor is directly proportional to the charge on it. Φ α V Q = CV Φ = Q/V where C is constant C: Capacitance or Capacity of capacitor. "capacity is also defined as the ratio of the charge to the potential." * sl unit of charge is C potential is V * S1 unit of capacity or capacitance = C/V or farad (f) Dimension = [L^-2 MT⁴A²] 1C 1 farad = 1V 1 farad:- A capacitor has a capacitance of one farad if the potential difference across it is 1V when 1 coulomb of charge is given to it. commonly used units of capacitance:- 1μF = 10^-6F 1nF = 10^-9F 1pF = 10^-12F ## Combination of Capacitors! ## a) Capacitors in series - When the no. of capacitors are connected one after another end to end so that the current flowing through each Capacitor is same. Then the combination is called as series combination as shown in fig. Let q be the constant charge on each condenser. Let V1, V2, V3 be the potential differences across 1st, 2nd & 3^rd condenser respectively. V be total P.D. across series combination 4 it is given by, V = V₁ +V₂ +V₃ Let, C_s: equivalent capacitance of combination, q: Total charge V: Total P.D. .. V= q/C_s - (3) From egn (2) & (3), we get, q/C_S = q/C₁ + q/C₂ + q/C₃ 1/C_S = 1/C₁ + 1/C₂ + 1/C₃ - (4) If n condensers are connected in series then equivalent capacitance is 1/C_S = 1/C₁ + 1/C₂ + 1/C₃ +………..+ 1/C_n - (5) If all capacitances are equal then 1/C_S = 1/C + 1/C + 1/C +……….+ 1/C - (n times) C_S = C/n or C_S = C/n ## b) Capacitors in Parallel: When the number of capacitors are connected between two common points so that the potential difference across each condenser is same, then the combination is called as Parallel combination as shown in fig. Let C1, C2 & C3 be capacitance of three condensers connected in parallel as shown in fig. V: constant potential difference across each condenser q₁,q₂,q₃: charges on 1st, 2nd & 3^rd condenser. q: Total charge : q = q₁ +q₂ +q₃ - (1) But q = CV : q₁= C₁V q₂= C₂V q₃= C₃V Putting these values in eq (1) we get, q = C₁V +C₂V+ C₃V q = V (C₁+C₂+C₃) Now, C_p: equivalent capacitance of parallel combination : q = C_pV Egn (2) becomes. C_pV = V(C₁+C₂+C₃) : C_p = C₁+C₂+C₃ - (3) If n condensers are connected in parallel then equivalent capacitance is C_p = C1+C2 + C₃ +...........+ C_n - (4) If all capacitance are equal C_p = C + C + C---... = nC C_p = nC ## • Capacitance of parallel plate capacitor:- ## a) Capacitance of a parallel plate capacitor without a dielectric. A parallel plate condenser or capacitor consists of two conducting plates P1&P2 parallel to each other. Let A be area of each plate d-distance between plates P₁&P₂ When charge q is given to insulated plate, then a charge -q is induced on the inner face of P₂ & free positive charge is induced on outer side of P₂ hence plate P₂ is earthed. In the outer regions, electric field due to charged plate cancel out, the net field is zero. E = σ/ε₀ - σ/ε₀ In the inner regions between the two capacitor plates, the electric field due to the two charge plates add up. The net field is E = σ/ε₀ + σ/ε₀ = 2σ/ε₀ We know, σ = q/A E = q/Aε₀ - (1) Let V be potential difference beth plates & it is given by, V = Ed = qd/Aε₀ But C = q/V - (3) .. egn (2) becomes C = q / (qd/Aε₀) C= (Aε₀)/d This is the expression for capacitance of parallel plate condenser. ## b) capacitance of a parallel plate capacitor with a dielectric slab between the plates:- Consider a parallel plate capacitor plate with two plates each of area A seperated by a distance d. :: Capacitance is given by, C₀= Aε₀/d - (1) Eo Electric field intensity between the plates before the introduction of dielectric slab. The potential difference between the plates is Where, V₀ = E₀d E₀= σ/ε₀ = Aσ/ε₀ σ= surface charge density on plate A dielectric slab of thickness t (t<d) is introduced between the plate of capacitor. The field E₀ polarizes dielectric, and induces charge -σ_p on left side & +σ_p on right side of dielectric as shown in fig. These induced charges set up a field Ep inside dielectric in opposite direction of E₀ & it is given by, Ep= σ_p/ε₀ = σ_p/Aε₀ The net field (E) inside dielectric reduces to (E₀-Ep) :: E = E₀- Ep E= E₀- E_p/E₀ ( as K = E₀/E₀-Ep) E = E₀/K Putting E₀ = q/Aε₀ E = q/AKε₀ .. q = EAKε₀ - (2) The field E exist over a distance t & E₀ over remaining distance (d-t) between the capacitor plates. Hence P.D. between the plate is V = E₀(d-t) + E(t) .. (E = E₀/K) V = E₀(d-t) + E₀(t)/k = E₀ [ (d-t) + t/k] V = Φ/Aε₀ * [ (d-t) + t/k] By def", C = q/V C = q/ {Φ/Aε₀ * [ (d-t) + t/k] } C = (Aε₀)/((d-t) + t/k) - (3) C = (Aε₀)/(d-t+t/k) ## Special Cases:- ### Case I:- If the dielectric fills up the entire space then t=d C= Akε₀/ d ( :: C₀ = Ad/ε₀) C= KC₀ K= C/C₀ ### Case II: If the capacitor filled with n dielectric slabs of thickness t₁, t₂, t₃, ...... t_n C = (Aε₀)/( t₁/k₁ + t₂/k₂ + .... + t_n/k_n) ### Case III- If the arrangement consist of n capacitor in parallel with plate areas A₁, A₂, A_3, A_n & plate separation d C= (ε₀( A₁k₁+A₂k₂+......A_nk_n))/d IF A₁ = A₂ = A₃ = .... =An C = Aε₀(k₁+k₂+....+k_n)/d ### Case IV. If the capacitor is filled with conducting slab (K=∞) then C = (Aε₀)/(d-t) (C)(∞) Thus, the capacitance is increased by the factor (d/d-t). # *Displacement Current:-* A parallel plate capacitor with dielectric is connected across D.C. Source as shown in fig. In conducting part of circuit, current flows but in region between plates of capacitors there are no free electrons available for conduction. In dielectric hence, current does not flow. Displacement current (I_d) When the circuit is closed, Current I_c flows through circuit in finite time and current I_c is found to be same everywhere in circuit except inside capacitors. As the current passes through the capacitors, the electric field between plates increases and it causes polarisation of dielectric, Thus, there is current in dielectric due to movement of bound charges. The current due to bound charges is called displacement current (I_d). The charge produced on plates of capacitor due to electric field E is given by, q = Aκε₀ E - (1) Differentiating eqn①, we get, dq/dt = Aκε₀ dE/dt where dq/dt = I_c = conducting current flowing in conducting part of circuit, .. eqn (2) becomes, I_c = Aκε₀ dE/dt dE/dt = I_c/Aκε₀ .. The rate of change of electric field (dE/dt) across capacitor is directly proportional to current I_c following in conducting part of circuit. From egn ②, Aκε₀ dE/dt is displacement current I_d is caused by displacement of bound charges in dielectric of capacitors under the influence of electric field. It have dimension of current. I_d = Aκε₀ dE/dt - (3) It means I_d = I_c For air or vacuum K=1 I_d = Aε₀ dE/dt A displacement current (I_d) exists at any point in space where, time varying electric field (E) exists. # Energy stored in capacitor:- A capacitor stores energy, when a capacitor is charged, it does some work as electrostatic energy is stored in the capacitor. Consider a capacitor of capacitance C being charged by DC source of volt V.