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Published byYvonne Braselton Modified over 2 years ago

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Electrostatic principles

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

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Charging a capacitor CR 2CR 3CR 4CR Charge Current Time Q = Q o (1-e -t/RC ) = CV b (1-e -t/RC ) I = I 0 e –t/RC = (V b /R) e –t/RC

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

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Charging a capacitor An RC network comprises an 6 μ F 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.

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Charging a capacitor i) V/R = 150/(0.4 x 10 6) = 375 μA ii) time constant = RC = (0.4 x 10 6) x (6 x 10 –6) = 2.4 Seconds

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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 Log n 0.33 = -t/2.4 -1.1 = -t/2.4 t = 1.1x2.4 = 2.64 secs

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Charging a capacitor iv) voltage after 2 seconds. v = 150 (1 - e - 2/2.4 ). 150 (1- 0.43) =150 x.56 84 volts

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Charging a capacitor current after 2seconds i = Ie - t/2.4. Therefore I= (375 x 10 –6) x 0.43) = 155 x 10 -6 A

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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 x10 -3 Amps 3.2mA

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

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

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

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Deflection of an electron beam ve Centre line of beam Cathode (-) Anodes (+) deflecting plates One eV = 1.6 x 10 -19 Joules (Charge on an electron = 1.6 x10 -19 coulombs)

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Deflection of an electron beam ve Centre line of beam

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Deflection of an electron beam ve Centre line of beam 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

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Deflection of an electron beam ve Centre line of beam Vertical deflection of electron From s = ut + 0.5at 2 (suvat equation) Length of plate (l)

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

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Deflection of an electron beam d = 0.5 x a x t 2 = 0.5 x (q V/dm) x l 2 /v 2

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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 10 6 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.

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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

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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 x10 15

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Deflection of an electron beam Example From s = ut + 0.5at 2 (suvat equation) d = 0.5 x a x t 2 = 0.5 x (q V/dm) x l 2 /v 2 0.5 (1.6x10 -19 x 60)/(12 x10 -3 x 9.1 x10 -31 ) X (24x10 -3 ) 2 /(15 X10 6 ) 2 1.12mm

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Electric field from a point charge : E = k Q / r 2 (k =9.0 × 10 9 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

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Other equations k = 1/4πε 0 (ε 0 = permittivity of free space) F = qE (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)

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