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1. Physics

2. ELECTROSTATICS - I

3. Session Objectives • Charges and their properties • Coulomb’s law • Introduction to electric field

4. Electric Charge - + Intrinsic property of matter by virtue of which it can (interact) exert force (electric) on other matter having the same intrinsic property Two types – positive and negative (arbitrarily assigned) SI unit of charge – Coulomb (C)

5. Methods of charging a body Frictional rubbing Transfer of electrons from one body to another cause charging Induction A charged body induces its opposite charge on the Nearby side of another body Conduction Charge on a body conducts to another

6. - - - + + + Properties of charge Like charges repel Unlike charges attract Hair stands out - each strand has the same electric charge and is being repelled from every other strand and her body

7. Properties of charge q 3q 4q + = 5q –7q –2q + = Conservation 2e e –3e 1.4e 3.5e Charge is conserved— Neither created nor destroyed, Can be just transferred Additivity Sign should be taken into account Total charge on a system – the algebraic sum of all charges contained in it. Quantisation Charge cannot be sub-divided indefinitely. The smallest unit of charge is equal to that of an electron I.e. e = 1.6 x 10-19 Other charges – integral multiple of e

8. Solved Example – 1 What kind of charges is produced on each, when (a) A glass rod is rubbed with silk? (b) An ebonite rod is rubbed with wool? Solution: (a) On glass rod – positive charge On silk – negative charge (electrons are transferred from glass to silk) (b) On ebonite rod – negative charge On wool – positive charge (Electrons are transferred from wool to ebonite)

9. Solved Example – 2 From a neutral body, 107electrons are removed. Find the charge the body acquires. Solution: Charge q = ne = 107 x 1.6 x 10–19 = 1.6 x 10–12 C Since the electrons have been removed, the body is positively charged.

10. Coulomb’s Law d q1q2 > 0 repulsive q1q2 < 0 attractive q1 q2 Absolute permittivity of vacuum in a medium of dielectric constant K Force between two charges or charged bodies F acts along the line joining the two charges When SI system is used, in air / vacuum

11. Coulomb’s Law q1q2 < 0 q1 q2 q1q2 > 0 q1 q2 Vector form Force on q1 by q2 Force on q2 by q1

12. Coulomb’s Law in vacuum Definition of coulomb Put q1 = q2 = 1 C and d = 1 m, then, One coulomb is that charge, which when placed at 1 m from an equal and similar charge in vacuum, repels it with a force of 9×109 N

13. Solved Example - 3 Why is Coulomb’s force a central force? Solution: Coulomb’s force between two charges always acts along the line joining them and hence, it is a central force.

14. Solved example – 4 Give the dissimilarities between Coulomb’s force and force of gravitation. Solution:

15. Addition of coulomb’s forces q1 q0 q2 q3 When a test charge is introduced to a system of charges, each of the charges of the system exert electric force on the test charge and the resultant force on it is the vector sum of the individual forces. Net force on q0 , In general,

16. Q q Q 2 x Solved Example – 5 A charge q is placed at the midpoint of the line joining two identical charges Q. What should be the value of q if this system of three charges is to be in equilibrium? Solution: For equilibrium,

17. Continuous charge distribution Does not mean that charge is continuous, but distribution of discrete charges is continuous A system of closely spaced charges form a continuous charge distribution Charge distribution Volume distribution of charge Linear charge distribution Surface charge distribution Force on q0

18. Electric field (intro.) A vector represented by A charge creates a physical condition in the space surrounding itself called the electric field such that, any other charge present in that space experience electric force. No boundary, but strength considerably decreases away from the charge responsible for that. Intensity of field at a point – Force on unit positive charge placed at that point. SI unit: N C–1 At a point, where strength of the filed is E, a charge q experiences a force of magnitude F = q E

19. Solved Example – 7 If an oil drop of weight 3.2 x 10–13 N is balanced in an electric field of 5 x 105 N C–1, then find the charge on the oil drop. Solution: W = 3.2×10–13 N, E = 5×105 N C–1 W = q E

20. Thank you