Physics

ELECTROSTATICS - 3

Session Objectives Application of Gauss’ law Electric potential (intro.)

Gaussian surface Any imaginary closed surface. Gaussian surfaces are useful in computing Electric field (usually uniform) and flux Gaussian surfaces are very useful for finding electric field with symmetrical charge distribution

Electric field due to an infinitely long straight charged wire Due to cylindrical symmetry of charge, electric field is away from the centre of the wire The net electric flux,

r + Charge enclosed by the cylinder, q = l h By Gauss’ law, 1. Gaussian surfaces should not pass through discrete charges, as electric field is not well defined in any location 2. Gaussian surfaces can easily pass through continuous charge distribution.

Solved Example – 1 An infinite line charge produces a field of 9×104 N C–1 at a distance of 2 cm. Calculate the linear charge density. Solution: E = 9×104 N C–1, r = 2×10–2 m

Solved Example –2 – + An electric dipole consists of charges +1.6 nC and –1.6 nC separated by a distance of 2×10–3 m. It is placed near a long line charge of linear charge density 5×10–4 Cm–1 as shown in the figure. If the negative charge is at a distance of 2 cm from the line charge, then, find the net force on the dipole – + 2 cm

Solution = 5×10–4 Cm–1, r1 = 2×10–2 m, r2 =2.2×10–2 m, q1 =-1.6×10–9 C, q2 =1.6×10–9 C Electric field due to a line charge at a distance r from it, Field at the point of negative charge, Force on the negative charge

Solution Contd. F1 is towards the line charge Electric field at the point of positive charge is – + F1 F2 Force on the positive charge F2 is away from the line charge Net force on the dipole, F =F1 – F2 = 0.72 – 0.65 =0.07 N, towards the line charge

Solved Example - 3 r A long cylindrical wire carries a positive charge of linear charge density 4×10–7 Cm–1. An electron revolves round the wire in a circular path under the influence of the electrostatic force. Find the KE of the electron. r Solution Let the electron revolves in a circular path of radius r. Electrostatic force on the electron provides the necessary centripetal force. Field at a distance r from the wire of charge,

Solution Contd. Force on the electron, Required centripetal force Kinetic energy of the electron,

Electric field due to infinite plane sheet of charge Thin, infinite, nonconducting sheet having uniform surface charge density, as a result of which surface of the sheet

Proof. At R and S, Electric flux at plane faces, Electric flux over the curved surface is zero, as no field lines crosses it Total electric flux,

Charge enclosed by the Gaussian surface, By Gauss’ law, E is independent of the distance from the plane sheet

Two infinite parallel sheets of charge Due to uniform surface charge density, surface of the sheet Region II Region I

Two infinite parallel sheets of charge Region III

Special Case The sheets have equal and opposite charge density If and , then I II III In regions I and III, E = 0 In region II,

Solved Example – 4 Two large thin metal plates with surface charge densities of opposite signs but equal magnitude of 44.27 ×10–20 Cm–2 are placed parallel and close to each other. What is the field To the left of the plates? (ii) To the right of the plates? (iii) Between the plates? Solution: Electric field exists only in the region between the plates. Therefore, (i), (ii) E = 0

Applying Gauss’ Law – Spherical Symmetry Electric field due a charged sphere of charge q and radius R (i) Field at an outside point, i.e., r > R Gaussian surface Note that E is same at all points on the Gaussian surface The field is the same as if the whole charge is placed at the centre of the shell

(ii) At a point on the surface, r = R (iii) At a point inside the spherical shell r < R Gaussian surface No charge is enclosed by the Gaussian surface.

r q + + r + + + + + + + + Distance of gaussian surface is from the centre. Radius incorrectly shown. Variation of electric field intensity from the centre of the shell with distance

Solved Example –5 The uniform surface charge density on a spherical copper shell is s. What is the electric field strength on the surface of the shell? Solution: The electric field on the surface of a uniformly charged spherical conductor is given by,

Solved Example –6 A spherical charged conductor has a uniform surface charge density s . The electric field on its surface is E. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ? Solution: The electric field on the surface of a uniformly charged spherical conductor is given by, Thus, E is independent of the radius of the sphere. As s is constant, E remains the same.

Solved Example –7 The uniform surface density of a spherical conductor is and the electric field on its surface is E1. The uniform surface density of an infinite cylindrical conductor is and the electric field on its surface is E2. Is the expression correct? Solution: The electric field on the surface of a charged spherical conductor is given by, (i)

Solution Contd. The electric field on the surface of a charged cylinder is given by, ( r = radius and l = length ) (ii) From equations (i) and (ii) we get,

Solved Example – 8 A spherical shell of radius 10 cm has a charge 2×10–6 C distributed uniformly over its surface. Find the electric field (a) Inside the shell (b) Just outside the shell (c) At a point 15 cm away from the centre Solution: q = 2 ×10–6 C, R = 0.1 m, r = 0.15 m (a) Inside the shell, electric field is zero

Solution Contd.

Electric potential Electric potential at a point - the work done in bringing unit positive charge from infinity to that point against the electric forces. SI unit – Volt (V) One volt - the electric potential at a point if one joule of work is done in bringing unit positive charge from infinity to that point.

Solved Example - 9 If a positive charge be moved against the electric field, then what will happen to the energy of the system? Solution: If a positive charge be moved against the electric field, then energy will be used from an outside source.

Solved Example -10 If 80 J of work is required to transfer 4 C charge from infinity to a point, find the potential at that point Solution: W =80 J, q = 4 C, V =?

Thank You