12th cbse unit 1 part 4

2. Line Integral of Electric Field (Work Done by Electric Field): Negative Line Integral of Electric Field represents the work done by the electric field on a unit positive charge in moving it from one point to another in the electric field. +q 0 O Z Y X + q Total work done by the electric field on the test charge in moving it from A to B in the electric field is F A r A B r B r W AB = dW = - E . d l A B Let q 0 be the test charge in place of the unit positive charge. The force F = +q 0 E acts on the test charge due to the source charge +q. It is radially outward and tends to accelerate the test charge. To prevent this acceleration, equal and opposite force –q 0 E has to be applied on the test charge. = 1 1 qq 0 4 π ε 0 ] [ - r B r A W AB = dW = - E . d l A B

4. Electric potential is a physical quantity which determines the flow of charges from one body to another. It is a physical quantity that determines the degree of electrification of a body. Electric Potential at a point in the electric field is defined as the work done in moving (without any acceleration) a unit positive charge from infinity to that point against the electrostatic force irrespective of the path followed. Electric Potential: According to definition, r A = ∞ and r B = r (where r is the distance from the source charge and the point of consideration) SI unit of electric potential is volt (V) or J C -1 or Nm C -1 . Electric potential at a point is one volt if one joule of work is done in moving one coulomb charge from infinity to that point in the electric field. or = 1 1 qq 0 4 π ε 0 ] [ - r B r A W AB = - E . d l A B = 1 1 q 4 π ε 0 ] [ - r B r A W AB q 0 = q 4 π ε 0 r W ∞ B q 0 = V V = W ∞ B q 0

6. Electric Potential due to a Single Point Charge: + q ∞ +q 0 Let +q 0 be the test charge placed at P at a distance x from the source charge +q. To prevent this acceleration, equal and opposite force –q 0 E has to be applied on the test charge. The force F = +q 0 E is radially outward and tends to accelerate the test charge. Work done to move q 0 from P to Q through ‘dx’ against q 0 E is dW = q 0 E dx cos 180 ° = - q 0 E dx or Total work done to move q 0 from A to B (from ∞ to r ) is P r B x Q dx q 0 E E dW = F . dx = q 0 E . dx dW = - dx q q 0 4 π ε 0 x 2 E = q 4 π ε 0 x 2 W ∞ B = ∞ B dW = - dx q q 0 4 π ε 0 x 2 ∞ r = - dx q q 0 4 π ε 0 x 2 ∞ r x 2 1 = q 4 π ε 0 r W ∞ B q 0 V = q 4 π ε 0 r

8. Electric Potential due to an Electric Dipole: + q - q A B +1 C V P = V P q+ + V P q- i) At a point on the axial line: l l x P p q V P q+ = 4 π ε 0 (x – l ) 1 V P q- = 4 π ε 0 (x + l ) 1 - q V P = 4 π ε 0 q (x – l ) 1 [ - (x + l ) 1 ] V P = 1 4 π ε 0 q . 2 l (x 2 – l 2 ) V P = 1 4 πε 0 p (x 2 – l 2 ) O

9. + q - q A B ii) At a point on the equatorial line: V Q = V P q+ + V P q- BQ = AQ The net electrostatic potential at a point in the electric field due to an electric dipole at any point on the equatorial line is zero. p θ θ y O q V Q q+ = 4 π ε 0 BQ 1 V Q q- = 4 π ε 0 AQ 1 - q V Q = 4 π ε 0 q BQ 1 [ - AQ 1 ] V Q = 0 Q l l

10. Equipotential Surfaces: A surface at every point of which the potential due to charge distribution is the same is called equipotential surface. i) For a uniform electric field: V 1 V 2 V 3 ii) For an isolated charge: Plane Equipotential Surfaces Spherical Equipotential Surfaces E E +

11. Properties of Equipotential Surfaces: 2. The electric field is always perpendicular to the element d l of the equipotential surface. 1. No work is done in moving a test charge from one point to another on an equipotential surface. If A and B are two points on the equipotential surface, then V B = V A . or Since no work is done on equipotential surface, i.e. E d l cos θ = 0 As E ≠ 0 and d l ≠ 0, cos θ = 0 or θ = 90 ° W A B q 0 V B - V A = ∆V = W A B q 0 = 0 W A B = 0 W AB = - E . d l A B = 0

13. Electrostatic Potential Energy: The work done in moving a charge q from infinity to a point in the field against the electric force is called electrostatic potential energy. W = q V i) Electrostatic Potential Energy of a Two Charges System: O Z Y X A (q 1 ) B (q 2 ) or r 1 r 2 - r 1 r 2 U = q 1 q 2 4 π ε 0 │ │ - r 1 r 2 1 U = q 1 q 2 4 π ε 0 r 12 1

14. ii) Electrostatic Potential Energy of a Three Charges System: O Z Y X A (q 1 ) B (q 2 ) or C (q 3 ) iii) Electrostatic Potential Energy of an n - Charges System: r 1 r 2 r 3 - r 1 r 3 - r 2 r 3 U = q 1 q 2 4 π ε 0 │ │ - r 1 r 2 1 + q 1 q 3 4 π ε 0 │ │ - r 1 r 3 1 + q 2 q 3 4 π ε 0 │ │ - r 2 r 3 1 U = q 1 q 2 4 π ε 0 r 12 1 [ q 1 q 3 r 31 q 2 q 3 r 32 + + ] 1 4 π ε 0 U = ∑ j=1 i ≠j n q i q j r j - r i │ │ ∑ i=1 n [ 2 1 ] - r 1 r 2

15. Area Vector: Small area of a surface can be represented by a vector. dS S Electric Flux: Electric flux linked with any surface is defined as the total number of electric lines of force that normally pass through that surface. 90 ° θ S Electric flux d Φ through a small area element dS due to an electric field E at an angle θ with dS is Total electric flux Φ over the whole surface S due to an electric field E is Electric flux is a scalar quantity. But it is a property of vector field. SI unit of electric flux is N m 2 C -1 or J m C -1 . dS n dS E dS θ θ dS dS = E dS cos θ E . dS d Φ = Φ = E . dS S = E S cos θ = E . S dS = dS n dS

17. Gauss’s Theorem: The surface integral of the electric field intensity over any closed hypothetical surface (called Gaussian surface) in free space is equal to 1 / ε 0 times the net charge enclosed within the surface. Proof of Gauss’s Theorem for Spherically Symmetric Surfaces: Here, O • r r +q E . dS = S Φ E = 1 ε 0 ∑ i=1 n q i E . dS d Φ = r 2 1 4 π ε 0 = q r . dS n d Φ = r 2 1 4 π ε 0 q dS r n . = 1 x 1 cos 0° = 1 r n . d Φ = r 2 1 4 π ε 0 q dS S Φ E = d Φ r 2 1 4 π ε 0 q = 4 π r 2 ε 0 q = dS S r 2 1 4 π ε 0 q = dS E

18. Proof of Gauss’s Theorem for a Closed Surface of any Shape: Here, r E E . dS d Φ = r 2 1 4 π ε 0 = q r . dS n d Φ = r 2 1 4 π ε 0 q dS r n . = 1 x 1 cos θ = cos θ r n . d Φ = r 2 q 4 π ε 0 dS cos θ S Φ E = d Φ q 4 π ε 0 = 4 π ε 0 q = d Ω q 4 π ε 0 S = d Ω θ dS n r +q •

19. O • r r +q Deduction of Coulomb’s Law from Gauss’s Theorem: From Gauss’s law, or If a charge q 0 is placed at a point where E is calculated, then which is Coulomb’s Law. or dS E E . dS = S Φ E = q ε 0 E dS = S Φ E = q ε 0 dS = S Φ E = q ε 0 E E = q 4 π ε 0 r 2 E x 4 π r 2 q ε 0 = Since E and dS are in the same direction, F = qq 0 4 π ε 0 r 2

20. Applications of Gauss’s Theorem: 1. Electric Field Intensity due to an Infinitely Long Straight Charged Wire: Gaussian surface is a closed surface, around a charge distribution, such that the electric field intensity has a single fixed value at every point on the surface. From Gauss’s law, = E x 2 π r l - ∞ + ∞ B A C E . dS = S Φ E = q ε 0 E . dS = S E . dS + A E . dS + B E . dS C E . dS = S E dS cos 90 ° + A B E dS cos 90 ° + C E dS cos 0 ° = C E dS E E E dS dS dS l r

21. (where λ is the liner charge density) or or In vector form, The direction of the electric field intensity is radially outward from the positive line charge. For negative line charge, it will be radially inward. Note: The electric field intensity is independent of the size of the Gaussian surface constructed. It depends only on the distance of point of consideration. i.e. the Gaussian surface should contain the point of consideration. q ε 0 = λ l ε 0 E x 2 π r l = λ l ε 0 E = 2 πε 0 1 λ r E = 4 πε 0 1 2 λ r E (r) = 4 πε 0 1 2 λ r r

22. C B A l 2. Electric Field Intensity due to an Infinitely Long, Thin Plane Sheet of Charge: From Gauss’s law, σ = 2E x π r 2 TIP: The field lines remain straight, parallel and uniformly spaced. E dS E E dS dS r E . dS = S Φ E = q ε 0 E . dS = S E . dS + A E . dS + B E . dS C E . dS = S E dS cos 0 ° + A B E dS cos 0 ° + C E dS cos 90 ° = 2E dS

23. (where σ is the surface charge density) or In vector form, The direction of the electric field intensity is normal to the plane and away from the positive charge distribution. For negative charge distribution, it will be towards the plane. Note: The electric field intensity is independent of the size of the Gaussian surface constructed. It neither depends on the distance of point of consideration nor the radius of the cylindrical surface. If the plane sheet is thick, then the charge distribution will be available on both the sides. So, the charge enclosed within the Gaussian surface will be twice as before. Therefore, the field will be twice. E = 2 ε 0 σ E ( l ) = 2 ε 0 σ l q ε 0 = σ π r 2 ε 0 2 E x π r 2 = σ π r 2 ε 0 E = ε 0 σ

24. 3. Electric Field Intensity due to Two Parallel, Infinitely Long, Thin Plane Sheet of Charge: σ 1 σ 2 Region I Region II Region III Case 1: σ 1 > σ 2 E 1 E 1 E 1 E 2 E 2 E 2 E E E E = E 1 + E 2 E = 2 ε 0 σ 1 + σ 2 E = E 1 - E 2 E = 2 ε 0 σ 1 - σ 2 E = E 1 + E 2 E = 2 ε 0 σ 1 + σ 2 σ 1 > σ 2 ( )

25. Case 2: σ 1 σ 2 Region I Region II Region III + σ 1 & - σ 2 E 1 E 1 E 1 E 2 E 2 E 2 E E E E = E 1 - E 2 E = 2 ε 0 σ 1 - σ 2 E = E 1 + E 2 E = 2 ε 0 σ 1 + σ 2 E = E 1 - E 2 E = 2 ε 0 σ 1 - σ 2 σ 1 > σ 2 ( ) σ 1 > σ 2 ( )

26. Case 3: σ 1 σ 2 E = 0 Region I Region II Region III E = 0 + σ & - σ E 1 E 1 E 1 E 2 E 2 E 2 E ≠ 0 E = E 1 - E 2 E = 2 ε 0 σ 1 - σ 2 = 0 E = E 1 - E 2 E = 2 ε 0 σ 1 - σ 2 = 0 E = E 1 + E 2 E = 2 ε 0 σ 1 + σ 2 = ε 0 σ

27. 4. Electric Field Intensity due to a Uniformed Charged This Spherical Shell: q From Gauss’s law, or or i) At a point P outside the shell: Since q = σ x 4 π R 2 , Electric field due to a uniformly charged thin spherical shell at a point outside the shell is such as if the whole charge were concentrated at the centre of the shell. HOLLOW ……… Gaussian Surface dS E r R • P E . dS = S Φ E = q ε 0 E dS = S Φ E = q ε 0 dS = S Φ E = q ε 0 E E x 4 π r 2 q ε 0 = Since E and dS are in the same direction, E = q 4 π ε 0 r 2 E = ε 0 r 2 σ R 2 O •

28. From Gauss’s law, or or ii) At a point A on the surface of the shell: Electric field due to a uniformly charged thin spherical shell at a point on the surface of the shell is maximum. Since q = σ x 4 π R 2 , dS E E . dS = S Φ E = q ε 0 E dS = S Φ E = q ε 0 dS = S Φ E = q ε 0 E E x 4 π R 2 q ε 0 = Since E and dS are in the same direction, E = q 4 π ε 0 R 2 E = ε 0 σ q R HOLLOW O • • A

29. From Gauss’s law, or or iii) At a point B inside the shell: This property E = 0 inside a cavity is used for electrostatic shielding. (since q = 0 inside the Gaussian surface) E = 0 r E R O E max END dS E r’ O q R HOLLOW • B • E . dS = S Φ E = q ε 0 E dS = S Φ E = q ε 0 dS = S Φ E = q ε 0 E E x 4 π r’ 2 q ε 0 = Since E and dS are in the same direction, E = 0 4 π ε 0 r’ 2