A parallel plate capacitor has a capacitance of C 1.0 nF and a charge of magnitu

Question

A parallel plate capacitor has a capacitance of C 1.0 nF and a charge of magnitude 0.20 nC. The plates are 1.0 mm apar a) What is the potential difference between plates? (0.20 V) b) What is the area of each plate? (0.11 m2) c) What is the electric field between the plates? (200 V/m) What is d) capacitance (2.1 nF) and e) maximum voltage which can be applied to that capacitor if the space between the parallel plates is filled with insulator with dielectric constant 2.1 and dialectic strength 60 x 10° V/m? (60 x 10' v) ( Two capacitors with Ci-5.0uP and C-2.0uF are connected to a potential difference of V-10.0V. a) Find the charge and potential difference for eachc much energy is stored into both capacitors? (350

Explanation / Answer

given

capacitance C = 1.0 nF = 1*10^-9 F

charge Q = 0.2 nC = 0.2*10^-9 C

seperation d = 1.0 mm = 1*10^-3 m

(a)

potential difference V = Q/C = 0.2 V

(b)

capacitacne C = eo*A/d

eo = permitivitty = 8.85*10^-12 C^2/Nm^2

A = area of plate

A = C*d/(eo)

A = 1*10^-9*1*10^-3/(8.85*10^-12)

A = 0.11 m^2

(c)

Electric field E = V/d = 0.2/(1*10^-3) = 200 V/m

(d)

capacitance C' = k*c = 2.1*1 nF = 2.1 nF

(e)

voltage V' = Emax*d = 60*10^6*1*10^-3 = 60*10^3 V

=====================================================

8)

In parallel potential each capacitor is same = 10 V

(a)

charge on C1 , Q1 = C1*V1 = 5*10 = 50 uC

potential difference across C1 , V1 = 10 V

charge on C2 , Q2 = C2*V1 = 2*10 = 50 uC

potential difference across C2 , V2 = 10 V

(b)

energy E1 = (1/2)*C1*V1^2 = (1/2)*5*10^-6*10^2 = 250 uJ

energy E2 = (1/2)*C2*V2^2 = (1/2)*2*10^-6*10^2 = 100 uJ

total energy stored E = E1 + E2 = 350 uJ

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