Electrostatic principles Field pattern in a capacitor Field

Electrostatic principles

Field pattern in a capacitor Field strength = V/d Volts per metre (voltage gradient)

Charging a capacitor Q = Qo(1 -e -t/RC) = CVb(1 -e -t/RC) Charge Current I = I 0 e –t/RC = (Vb/R) e –t/RC CR 3 CR 2 CR Time 4 CR

Discharging a capacitor Q = Qoe-t/CR Charge CR 3 CR 2 CR Time 4 CR

Charging a capacitor An RC network comprises an 6 μF capacitor and a 0. 4 MΩ resistor. When a 150 V d. c. supply is applied to the network, calculate: i) initial charging current ii) time constant iii) time taken for the p. d. across the capacitor to reach 100 V iv) current, and p. d. across the capacitor, 2 seconds after connecting the supply.

Charging a capacitor i) V/R = 150/(0. 4 x 106) = 375 μA ii) time constant = RC = (0. 4 x 106) x (6 x 10– 6) = 2. 4 Seconds

Charging a capacitor iii) v = V (1 -e -t/RC ). 100 = 150 (1 - e -t/2. 4). (100/150) - 1 = -e -t/2. 4 = -0. 33 e -t/2. 4 = 0. 33 Logn 0. 33 = -t/2. 4 -1. 1 = -t/2. 4 t = 1. 1 x 2. 4 = 2. 64 secs

Charging a capacitor • iv) voltage after 2 seconds. • v = 150 (1 - e- 2/2. 4). • 150 (1 - 0. 43) • =150 x. 56 • 84 volts

Charging a capacitor • current after 2 seconds i = Ie - t/2. 4. • Therefore I= (375 x 10– 6) x 0. 43) • = 155 x 10 -6 A

Instantaneous current • Instantaneous current, • I = C x dv/dt • If the voltage across a 4μF is changing at the rate 800 v/s the current flow is • 4 x 10 -6 x 800 • = 3. 2 x 10 -3 Amps • 3. 2 m. A

Deflection of an electron beam Anodes (+) ve ve Cathode (-) Centre line of beam deflecting plates Electrons are accelerated from the cathode through the anodes

Deflection of an electron beam Anodes (+) ve ve Cathode (-) Centre line of beam deflecting plates Electrons are accelerated from the cathode through the anodes

Deflection of an electron beam Anodes (+) ve ve Cathode (-) Centre line of beam deflecting plates An electron accelerated through a potential difference of 1 volt gains an electron-volt (e. V) of energy

Deflection of an electron beam Anodes (+) ve ve Cathode (-) Centre line of beam deflecting plates One e. V = 1. 6 x 10 -19 Joules (Charge on an electron = 1. 6 x 10 -19 coulombs)

Deflection of an electron beam ve Centre line of beam ve

Deflection of an electron beam ve Centre line of beam ve Charge on an electron q = 1. 6 x 10 -19 coulombs Force acting on the electron = q x potential gradient F = q x v/d

Deflection of an electron beam ve Centre line of beam ve Length of plate (l) Vertical deflection of electron From s = ut + 0. 5 at 2 (suvat equation)

Deflection of an electron beam • ut =0 • (initial velocity indirection of deflection = 0) • a = F/m • =(q. V/d) ÷ m • q V/dm • t = l/v • (l = length of plate, v = electron velocity)

Deflection of an electron beam d = 0. 5 x a x t 2 = 0. 5 x (q V/dm) x l 2/v 2

Deflection of an electron beam Example • If the axial length of the plates is 24 mm and spacing 12 mm apart, and the axial velocity of the beam entering the plates is 15 x 106 m/s. Calculate, for a 60 V deflecting voltage, the transverse distance travelled by the electron beam at the point of exit from the deflecting plates. • •

Deflection of an electron beam Example • The field strength = V/d 60/0. 12 = 5000 volts/m • Calculate the force exerted on the electron beam as it passes through the plates • = q x Potential gradient • = 1. 6 x 10 -19 x 5000 • = 8 x 10 -16 Newtons •

Deflection of an electron beam Example • The acceleration of the diverted beam resulting from the force applied • = force /mass • 8 x 10 -16/9. 1 x 10 -31 • 0. 88 x 1015

Deflection of an electron beam Example • From s = ut + 0. 5 at 2 • (suvat equation) • d = 0. 5 x a x t 2 • = 0. 5 x (q V/dm) x l 2/v 2 • 0. 5 (1. 6 x 10 -19 x 60)/(12 x 10 -3 x 9. 1 x 10 -31) • X • (24 x 10 -3)2/(15 X 106)2 • 1. 12 mm

Electric field from a point charge : E = k Q / r 2 (k =9. 0 × 109 N m 2/C 2 ) E r E is the field strength Q is the charge and r is the distance from the centre of the circle

Other equations k = 1/4πε 0 (ε 0 = permittivity of free space) F = q. E (F = q x v/d) (force = charge x field strength) Electric flux density D = Q/A [coulombs/metre 2] Force between 2 charges (+ if unlike charges) (- if like charges)