Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Anti-gravity wikipedia , lookup
Work (physics) wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Speed of gravity wikipedia , lookup
Magnetic monopole wikipedia , lookup
History of electromagnetic theory wikipedia , lookup
Circular dichroism wikipedia , lookup
Electromagnetism wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Maxwell's equations wikipedia , lookup
Field (physics) wikipedia , lookup
Lorentz force wikipedia , lookup
Physics ELECTROSTATICS - 2 Session Objectives Electric field due to a point charge Electric field due to a dipole Torque and potential energy of a dipole Electric lines of force Gauss’ Law Electric field due to a point charge The test charge q0 is placed at M. The force on it, 1 qq0 F= d 3 4ε0 d Electric field strength at M, E= Or, F 1 q = d 3 q0 4ε0 d E= 1 q 4ε0 d2 The direction of electric filed is away from the positive point charge Solved Example - 1 Find the electric field due to a point charge of 0.5 mC at a distance of 4 cm from it in vacuum. How does the filed strength change if the charge is in a medium of relative permittivity 80? Solution: q =0.5×10–3 C, d = 4×10–2 m, k = 80 1 q 0.5×10–3 9 10 –1 (i) E= =9×10 × =2.82×10 NC 2 4 ε0 d2 4×10–2 1 q 2.82×1010 8 –1 (ii) E'= = =3.53×10 NC 4 ε0k d2 80 Superposition Principle Electric field due to a number of charges at a point is the vector sum of the fields due to individual charges. Rules of vector addition, as is the case for electric forces, is to be applied. Solved Example –2 Two point charges +q and +4q are separated by a distance 6a. Find the point on the line joining the charges where the electric field is zero. Solution: The field due to the charges should be equal and opposite at this point. Let the point be at a distance x from the charge +q. Then, 1 q 1 4q = x = 2a 4ε0 x2 4ε0 (6a– x)2 The point is at a distance 2a from the charge +q +q +4q x 6a–x Electric Dipole A system of two equal and opposite charges separated by a certain distance Dipole moment - a measure of the strength of electric dipole. It is a vector quantity represented by P . Magnitude of dipole moment - product of the magnitude of either charge and the separation between them. Direction of dipole moment – it points from negative towards positive charge P = q (2a) SI unit – C m Solved Example - 3 Two charges 20 mC and – 20 mC are 4 mm apart. Determine the electric dipole moment of the system. If the charges are in a medium, will the dipole moment change? Solution: q = 20 ×10–6 C, 2a = 4×10–3 m (i) p = q(2a) = 20×10–6×4×10–3 = 8 ×10–8 C m (ii) Dipole moment is independent of the nature of the medium between the charges Electric field due to a dipole (i) Along the axial line The resultant field at M, E = E1+E2 E= q 2 4ε0 d+a ˆ q .n = 4ε0 E= 2p 4ε0 d3 ˆ ×(–n)+ 2d(2a) 2 2 2 d -a q 2 4ε0 d– a ×nˆ = 2p 3 4ε0 d ˆ ;as p =q(2a) n and d >> a Electric field due to a dipole (ii) On the equatorial plan E1 = E2 = q 4ε0 (d2 + a2 ) q 4ε0 (d2 + a2 ) E Sine components of E1 and E2 cancel out. Cos components add up E = E1 cosθ+E2 cosθ G E = = 2q 2 2 4ε0 (d + a ) cosθ q×2a 4ε0 d2 + a2 3/2 as cosθ = a d2 +a2 p = 3 4ε0d as p = q(2a) and d>> a The direction of electric field is opposite to that of dipole moment Therefore, E = p 4ε0 d3 Torque on a dipole In an electric field E, the force acting on charges will be equal and opposite. The parallel forces form a couple. Moment of the couple constitute a torque t= = = = force x perpendicular distance qE x AC qE x 2a sinq PE sinq – qE τ = p× E E A qE q B ;where, p is the dipole moment C Solved Example – 4 An electric dipole, when held at 30° with respect to a uniform electric field of 104 NC–1, experiences a torque of 9 ×10–26 Nm. Find its dipole moment. Solution: t = 9×10–26 Nm, q =30°, E =104 NC–1 t = PEsinθ τ 9×10–26 P= = 4 =1.8×10–29 Cm Esinθ 10 ×sin30° Potential Energy of a dipole Work done on a dipole in an electric field is stored as its potential energy Work done in rotating a dipole from orientation q1 to q2 W q2 q1 q2 q1 t(q) dq pE sin q dq q –pE cos qq2 1 =-pEcosθ2 - cosθ1 If q1 = 90° and q2 = q, then, W=-pE cosθ- cos90o =-pEcosθ Or, Potential energy of the dipole, U=–pE cosθ =–p.E Solved Example – 5 What does the negative sign in the expression for the potential energy of a dipole indicate? ( U=–p.E ) Solution: The potential energy of the dipole is taken to be zero, When it is oriented perpendicular to the direction of the electric field. It is the maximum potential energy, the dipole can acquire. In any other orientation, it is less than zero and hence, the negative sign. Electric line of force Imaginary curve drawn in an electric field whose tangent at any point gives the the direction of intensity of the field at that point. Lines of force originate at a positive charge and terminate at a negative charge. Tangent at any point to line of force shows the direction of the field at that point. Lines of force do not intersect and are continuous curves. Number density of lines of force at point gives the magnitude of electric field. Electric Field lines Point Charge +ve : radial, outward directed st. lines end at -ve : radial, inward directed st. lines begin at Both are uniformly distributed. No. of lines for charge |q| = q/ Electric Field lines 2. Two Positive Charges M is the neutral point Equal charges: same number density When the charges are unequal, M shifts towards the smaller charge. Electric Field lines 3. Two unlike Charges Unequal charges Equal charges: same number density –Ve > + Ve Electric Field lines Uniform Field From infinity Equispaced parallel straight lines To infinity Solved Example - 6 Why cannot two electric lines of force intersect each other? Solution: The tangent at a point on the line of force gives the direction of the field at that point. If two lines of force intersect at a point, the field will have two directions at that point, which is impossible. Electric Flux Electric flux over a surface is equivalent to the number of electric field lines passing normally through it. Denoted by f Mathematically, f E.ds Integration is carried over the surface and ds is a small area element. The direction of ds is perpendicular to the surface element SI unit of f : N m2 C–1. Gauss’ Law It relates electric fields at points on a closed surface and the net charge enclosed by that surface. The flux over a closed surface is 1 times charge enclosed by that surface ε0 qenc f= ε0 But ds f = E.ds s Therefore, ε0 E.ds = qenc s qenc + + - + + - q E Gauss’ Law If no charge is enclosed by the surface, fnet = 0 f q in f out Gauss law is useful in computing electric fields if the charge distribution is symmetric Solved Example - 7 The net outward flux through the surface of a black box is 8 ×103 Nm2C–1. (a) What is the net charge inside the box? (b) If the net outward flux were zero, could you say that there were no charges inside the box? Solution: (a) q = 8 ×103 Nm2C–1 If the net charge inside is q, then, q= q q = Φ ε0 = 8×103 ×8.854×10–12 = 7.08×10–8 C ε0 (b) Not necessarily. We can only say that the net charge inside the box is zero. Thank You