PLEASE ANSWER BOTH I NEED TO LEARN HOW TO SOLVE THESE TWO EQUATIONS. 1.What curr

Question

PLEASE ANSWER BOTH I NEED TO LEARN HOW TO SOLVE THESE TWO EQUATIONS.

1.What current flows through a 2.49 cm diameter rod of pure silicon that is 18.5 cm long, when 1000 V is applied to it? (Such a rod may be used to make nuclear particle detectors, for example.)
___A ?

2.

Figure 18.24

(a) Find the charge stored on each capacitor in Figure 18.24 (C1 = 30.0 µF, C2 = 4.50 µF) when a 1.51 V battery is connected to the combination.
C1
  ____C ?
C2
___C ?
0.300 µF capacitor
___C ?

(b) What energy is stored in each capactitor?
C1
___J ?
C2
___J ?
0.300 µF capacitor
___J ?

THANK YOU SO MUCH

Explanation / Answer

1) need resistivity of silicon

2)

let C3 = 0.3 micro F

Cnet = (C1+C2)*C3/(C1+C2+C3)

= (30+4.5)*0.3/(30+4.5+0.3)

= 0.297 micro F

charge on C3(0.3 micro F) = Cnet*V

= 0.297*10^-6*1.51

= 4.49*10^-7 F

potentila cross C1 and C2 = 1.51 - Q3/C3

= 1.51 - 0.449/0.3

= 0.01333 volts

charge on C1 = C1*V1

= 30*10^-6*0.0133

= 4*10^-7 C

charge on C2 = C2*V2

= 4.5*10^-6*0.0133

= 6*10^-8 C

b) Energy stored in C1 = Q1^2/(2*C1)

= (4*10^-7)^2/(2*30*10^-6)

= 2.67*10^-9 J

Energy stored in C2 = Q2^2/(2*C2)

= (6*10^-8)^2/(2*4.5*10^-6)

= 4*10^-10 J

Energy stored in C3 = Q3^2/(2*C3)

= (4.49*10^-7)^2/(2*0.3*10^-6)

= 3.36*10^-7 J

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.