Electrostatics

1. Coulomb’s Law r q2 q1 Definition : The force of attraction or repulsion between the two charges is directly proportional to the product of the magnitude of the two charges and inversely proportional to the square of distance between them i.e. εo = 8.85 × 10-12 C2/N-m2

2. Coulomb’s law ……….

3. F21 F12 F21 = F12 = F21 = F12 = F21 F12 = |F21| r12 r12 r12 r12 r21 r21 r21 r21 1 1 1 1 4peo 4peo 4peo 4peo q1q2 q1q2 q1q2 q1q2 ^ ^ ^ ^ r12 r21 r12 r21 r2 r2 r3 r3 _ |F12 | = = Coulomb’s Law …… Vector form : for like charges q2 q1 r + + Force is repulsive Forces are equal in magnitude and opposite in direction = r = r _ Since

4. F21 F12 F21 = F12 = F21 = F12 = F21 F12 = |F21| r12 r12 r12 r21 r21 r21 1 1 1 1 4peo 4peo 4peo 4peo q1q2 q1q2 q1q2 q1q2 ^ ^ r12 r21 r2 r2 r3 r3 _ |F12 | = = Coulomb’s Law …… Vector form : for unlike charges q2 q1 r - + Force is attractive Forces are equal in magnitude and opposite in direction _ Since Simulation

5. Superposition Principle F31 = F32 = F3 r13 F31 F32 r23 1 1 F3 = F31 + F32 Fi = Fij 4peo 4peo ^ ^ r23 r13 q1q3 q2q3 2 2 r13 r23 q3 + + q1 Resultant force on charge q3 q2 + This force can be obtained by parallelogram method In general, for group of ‘n’ point charges, the force on ith charge due to all other charges will be j=n  J=1 i≠j

6. Superposition Principle…….. F31 = F32 = r13 F31 F32 r23 1 1 F3 = F31 + F32 Fi = Fij 4peo 4peo ^ ^ r32 r31 F3 q1q3 q2q3 2 2 r13 r23 q3 - + q1 Resultant force on charge q3 q2 + This force can be obtained by parallelogram method Definition: Principle of superposition states that all the charges when placed near each other, behave independent of each other and the net force on any one of them is equal to the vector sum of the forces exerted on it due to all other charges. j=n  J=1 i≠j

7. Superposition Principle…….. F = F12 + F22 + 2F1F2 cos 90 F = F12 + F22 = 2 F1 F = 1.42 × 180 = 255.6 N F = F12 + F22 + 2F1F2 cos 90 F = F12 + F22 +2F1F2 cos 45 = 2 + 2 F1 F = 1.85 × 180 = 332.88 N F1 F1 F1 F F2 F F2 F = F2 Magnitude of resultant force is F = F12 + F22 + 2F1F2 cos θ - 2µC Example -  = 900  = 450 r = 0.01m - 2µC - r r θ θ - - + + r r - 2µC 1µC 1µC - 2µC In both the cases + F1 = 180 N andF2 = 180N In above cases Case I : θ = 900 Case II : θ = 450 Resultant force in second case is greater than the first case simulation

8. Electric Field F E qo F qo Electric Field An electric charge produces an electric field in the surrounding space When another charge is brought in this region of space, it experiences an electric force It is not charge which exerts the force but it is the field of a charge due to which another charge experiences an electric force Thus an electric field said to be exist in the region if another charge experiences an electric force in that region Electric Intensity : The intensity of an electric field at any point in an electric field is defined as the force acting on a unit positive charge placed at that point If is force acting at a point on test charge qo at a point, then electric intensity at that point is = The magnitude is E = The direction of electric intensity is along the direction of force acting on positive test charge. It’s unit is N/C

9. Electric Intensity ….. r r E = E E = 1 1 1 r 4peo 4peo 4peo ^ ^ ^ ^ r r r r q q q r2 r3 r2 r r Electric intensity due to point charge + P q The electric intensity is given by = Therefore, is unit vector along The magnitude is E =

10. Electric field….. Lines of force due to point charges - + For positive charge electric field is radially outwards For negative charge electric field is radially inwards simulation

11. Electric Field …… E3 E3 E2 E2 E1 E1 r1 E E r2 r3 Electric field due to group of point charges P The resultant electric intensity is q1 + = + + + q2 + q3 The electric intensity at any point is the vector sum of the electric intensities due to all the charges at that point

12. Electric Potential w qo 1 joule 1 coulomb The electric potential at any point in an electric field is defined as the work done in moving a unit positive charge from infinity to that point against the direction of electric field If W is work done in moving a test charge qo from infinity to a point, then potential at that point will be V = If W is in joule, qo is in coulomb, potential is involt 1 volt = The potential at a point in an electric field is called 1 volt when 1 joule work is done in moving 1 coulomb charge from infinity to that point

13. wA wB wB wB qo qo qo qo wAB qo Electric Potential ……. Potential difference : If WA is the work done by an external force in moving a positive test charge qo from infinity to point A, the potential at point A will be VA = If WB is the work done by an external force in moving a positive test charge qo from infinity to point B, the potential at point B will be VB = The potential difference between points A and B is VB - VA = = - The potential difference between two points is numerically equal to the work done in moving a unit positive charge from one point to another

14. Electric Potential…. E dW = F · dl The force must beF = - qoE dW = - qoE · dl dW dl - E · dl qo B B ∫ ∫ - - E · dl E · dl A ∞ Consider electric field shown by lines of force dl is small length element of the path B Work done in moving charge qo through dl by the application of external force is dl A dV = = The potential difference over the length dl is The potential difference between points A and B VAB = If point A is at infinity VB = This expression gives potential at point B simulation

15. r E - E · dx q q q 1 1 1 4peo 4peo 4peo x2 x2 r Electric potential due to a point charge B A P q + dx x Work done to displace unit positive charge through small distance dx against the field direction is = E dx dW = At point A, E = dx dW = On integration from x = ∞ to x = r, we get V =

16. q1 q2 1 1 4peo 4peo r1 r2 Potential at a point due to group of charges q1 q1 q1 + + + r1 r1 r1 + q2 r2 + r2 q2 P P q2 r2 P + Resultant potential at point P V1 + V2 V = + V = Note : Since potential is a scalar quantity, the potentials at point P will be the same in all above cases

17. Electric dipole p p = qd Two equal and opposite charges separated by a small distance form an electric dipole Equatorial line + q - q - + O Axial line d The product of the magnitude of either charge and the distance between the charges is called electric dipole moment Electric dipole moment : p = qd Unit of dipole moment is coulomb – metre or C-m It is directed along the axis of dipole and from – q to + q

18. Potential due to dipole p q q 1 1 1 4peo 4peo 4peo r1 r2 p cos θ r2 P r2 r1 r θ + q - q - + O _ d V = We get resultant potential V = wherep = qd

19. p p r2 p 1 4peo Electric Potential at axial point and equatorial point + q - q r P - + O θ = 00, cos 00 = 1 V = P r + q - q 900 - + O θ = 900, cos 900 = 0 V = 0

20. Electric Field due to dipole p 1 1 1 4peo 4peo 4peo E = Er2 + Eθ2 2p cos θ p sin θ p r3 r3 3 cos2θ + 1 r3 α = tan – 1(tan θ) 1 2 E Er Eθ α θ P Er = r2 r1 r Eθ = θ + q - q The resultant electric intensity is - + O The angle made by resultant electric field with the dipole moment is θ + α d E = The angle made by resultant electric field with the radial component of electric field is

21. E 2p p r3 1 1 4peo 4peo p 3 cos2θ + 1 r3 α = tan – 1(tan θ) 1 2 Electric Field at axial point + q - q r - + O P E = θ = 00, cos 00 = 1 Ea = We have, θ = 00, tan 00 = 0 α = tan – 1(0) = 0 Thus angle made by E with the dipole moment is θ + α = 0

22. E p p r3 1 1 4peo 4peo p 3 cos2θ + 1 r3 Electric field at equatorial point E = P r + q - q 900 - + O Ee = θ = 900, cos 900 = 0 θ = 900, tan 900 = ∞ α = tan – 1(∞) = 900 Thus angle made by E with the dipole moment is θ + α = 1800 Ea : Ee-> 2 : 1 Simulation

23. Electric dipole in electric field simulation

24. Gauss’s Theorem dS E q q εo εo S S ∫ ∫ E · dS E · dS Statement : The total electric flux passing through any closed surface is equal to q/εo The flux through the closed surface is  =  = q1 q3 = q2 q5 q4 S

25. Examples …… q εo q q  = 0 Because charge is outside the surface Total incoming flux is equal to total outgoing flux.  =

26. simulation Simulation 1