exercises are to be completed in full Exercises are worth 2 points ea.: 1 point

Question

exercises are to be completed in full

Exercises are worth 2 points ea.: 1 point for an attempt and 1 point for a correct answer. A capacitor is being designed such that when 10 V is applied 2 mC of charge separates out. The capacitor is a parallel plate type, and the separation of the plates is 0.0001 m. If the plates are square, and the material between the plates is vacuum, what is the length of a side of one plate? A 10 pF capacitor is constructed by placing two metallic sheets on top of each other, inserting a dielectric material that is 0.05 mm thick and has dielectric constant k = 5. The assembly is then folded up into a compact package. What is the area of the metallic sheets? Consider 1000 10 mu F capacitors, a) If they are all wired in series and then connected with 10V across the assembly, how much energy is stored? If they are all wired in parallel and then connected with 10V across the assembly, how much energy is stored? What is the resistance of a 10 cm length of aluminum wire that is 0.01 nun in diameter? How much current will flow through the wire from question 4 if the ends are connected across the terminals of a 9V batter? How much energy will be dissipated in 120 s?

Explanation / Answer

1. we know that

C = e*A/d where e is the dielectric constant, A is the area of the plate and d is the distance between the plate.

or,

Q/V = e*x^2/d , x is the length of the square plates.

putting the given value where e = 8.854*10^-12 we get,

x=47.5275 m ----------------------------ans.

2.in the same above formula use e = k*e0 where e0 = 8.854*10^-12.

so, we have,

10*10^-12 = 8.854*10^-12 *5*A/(0.05*10^-3) = 11.29433* 10^-6 square meter--------------------------ans.

3.

equivalent capacitance value can be found out as,

for n parallel cap C then,

Ceq = C/n------------------------a

and for parallel

Ceq = n*C---------------------b. so it is just opposite of the equivalent resistance.

so,

when they are connected in series Ceq = 10^-6/1000 = 10^-9 F

if they are connected in parallel then Ceq = 1000*10^-6 = 1mF

as we know that

E = 0.5*Ceq * V^2

so,

a) Ese = 0.5*10^-9 *100 = 50 *10^-9 J----------------------ans.

b) Epa = 0.5 *10^-3 *100 = 50 *10^-3 J-------------------------ans.

4. we know that, resistance is given by

R = r*L/A where r is resistivity which is material property and constant. L is the length and A is the area.

for the given problem A = pi*(0.01*10^-3/2)^2 square meter. and L = 0.1 m and r = 2.7*10^-8 ohm/m.

so,

R = 34.377 ohm.----------------------------ans.

5.

current is given by,

I = V/R = 9/34.377 A = 0.261799 A---------------ans.

enerhy is given by,

E = V*I*t , whre t is the time.

= 9 * 0.261799*120 = 282.7433 J -------------------------ans.

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