Capacitors 1. When the electric field in a certain dielectric exceeds 10 V/m it

Question

Capacitors 1. When the electric field in a certain dielectric exceeds 10 V/m it experiences dielectric breakdoun (gets destroyed). What is the breakdown voltage of a 10 x Ida. I F capacitor if K = 3.7? 2. In the circuit below V-= 12 volt, G = C,-1 nF. (a) Find C,9, Vi, ½,91-b and charge taken from the battery (b) find the total energy stored in the circuit (use C) (c) Finol C,, , Vi, ½, , 92, if the space between the lates of G (lower) is filled with K = 2. 3. In the circuit below 12 volt, G = G = 1 nF. ++ (a) Find C,q, li , ½ , 92 and charge taken from the battery (b) the same, if the space between the plates of Ca (right) is filled with 2. 4. In the circuit below V = 0 volt, G = G = G = 1 F. (a) Find C. and charge taken from the battery (b) Find V, V. a.. 9a s

Explanation / Answer

2) Given,

V = 12 V ; C1 = C2 = 1 n F

a)The equivalent capacitance will be:

Ceq = C1 + C2

Ceq = 1 + 1 = 2 n F

Since they are connected in parallel,

V1 = V2 = 12 V

q1 = C1 V1 = 1 x 12 = 12 n C

q2 = C2 V2 = 1 x 12 = 12 n C

Hence, Ceq = 2 n F , V1 = 12 V , V2 = 12 V , q1 = 12 n C ; q2 = 12 n C

b)U = 1/2 C V^2

U = 0.5 x 2 x 10^-9 x 12^2 = 1.44 x 10^-7 J

Hence, U = 1.44 x 10^-7 J

c)Now C2 = 2 n F since ; C = k e0 A/d

Equivalent capacitsnac will be:

Ceq = 1 + 2 = 3 nF

Since they are in parallel, drop will be same and

V1 = V2 = 12 V

q1 = C1V1 = 12 nC ;

q2 = C2V2 = 2 x 12 = 24 nC

Hence, Ceq = 3 nF ; V1 = V2 = 12 V ; q1 = 12 nC ; q2 = 24 nC.

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