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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 q 0 is placed at M. The force on it, .
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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, Electric field strength at M, Or, 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
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.
+q +4q x 6a–x 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, The point is at a distance 2a from the charge +q
Dipole moment - a measure of the strength of electric dipole. It is a vector quantity represented by . Electric Dipole A system of two equal and opposite charges separated by a certain distance 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,
G Sine components of cancel out. Cos components add up Electric field due to a dipole (ii) On the equatorial plan
The direction of electric field is opposite to that of dipole moment Therefore,
A 2a q B C 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
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
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 If q1 = 90° and q2 = q, then, Or, Potential energy of the dipole,
What does the negative sign in the expression for the potential energy of a dipole indicate? ( ) Solved Example – 5 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, inward directed st. lines begin at +ve : radial, outward directed st. lines end 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.
Unequal charges –Ve > + Ve Equal charges: same number density Electric Field lines 3. Two unlike Charges
From infinity To infinity Electric Field lines Uniform Field Equispaced parallel straight lines
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.
Mathematically, Integration is carried over the surface and is a small area element. The direction of is perpendicular to the surface element Electric Flux Electric flux over a surface is equivalent to the number of electric field lines passing normally through it. Denoted by f SI unit of f : N m2 C–1.
The flux over a closed surface is times charge enclosed by that surface qenc - + + - - - + - + - Gauss’ Law It relates electric fields at points on a closed surface and the net charge enclosed by that surface.
in out q Gauss’ Law If no charge is enclosed by the surface, fnet = 0 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, (b) Not necessarily. We can only say that the net charge inside the box is zero.
