A PARALLEL PLATE CAPACITOR HAS EACH PLATE WITH A SURFACE AREA OF 14.5cm^2, SEPAR

Question

A PARALLEL PLATE CAPACITOR HAS EACH PLATE WITH A SURFACE AREA OF 14.5cm^2, SEPARATED BY A DISTANCE OF 0 420mm. VOLUME BETWEEN THE PLATE IS VACCUM 1/3 A BATTERY OF EMF 240 V IS CONNECTED ACROSS THE CAPACTITOR PLATES. WHAT IS THE CAPACITANCE 1/3 CHARGE EACH PLATE? WHAT IS TOTAL ELECTRIC FIELD BETWEEN THE PLATES? WHAT IS TOTAL ENERGY STORED IN CAPACITOR WHEN BATTERY IS DISCONNECTED 1/3 Space in BETWEEN THE PLATE IS FILLED WITH DIELECTRIC OF LAMBDA = 10.00; WHAT IS THE NEW WNERGY STORED IN CAPACITOR?

Explanation / Answer

Capacitor surface area A = 14.5 cm2 = 14.5*10^-4 m2
separation of the plates d = 420 mm = 0.42 m
Battery of emf E = 240 v

a) capacitance of the capacitor is C = epsilon not A/d , here epsilon not is permitivity of free space
               = 8.85*10^-12*14.5*10^-4 / 0.42
               = 3.05*10^-14 F

charge on each plate is Q = C*V = 3.05*10^-14 *240 = 7.32*10^-12 C

b) electric field between the plates E = sigma/epsilon not

   sigma is surface charge density = q/A = 7.32*10^-12/14.5*10^-4 = 5.05*10^-19 C/m2

       E = 5.05*10^-19/8.85*10^-12 = 5.70*10^-8 N/C
c) total energy stored is U = 1/2 cV^2 = 1/2 *3.05*10^-14*240^2= 8.784*10^-10 J

d) if dielectric material placed between the plates then capacitance increaases by dielectric constant times

   U = 1/2 kC*v^2 = 1/2*10*3.05*10^-14*240^2 = 8.784*10^-9 J

increased by 10 times

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