could you show me your work please. A capacitor C_1 = 4.00 mu F is charged up to

Question

could you show me your work please.

A capacitor C_1 = 4.00 mu F is charged up to Delta V = 100 Volt. Another capacitor C_2 = 4.00 mu F has no charge (see picture 5.1 below). Without discharging, C_1 is connected to C_2 (picture 5.2 below). What is the chare q on C_1 before both capacitors are connected? What is the energy U stored on C_1 before both capacitors are connected? What is the total charge q' = q_1 + q_2 after two capacitors connected? d) What is the new voltage Delta V' across both capacitors after the two are connected?

Explanation / Answer

a) Q = cV = 4 x 10^-6 (100) = 4 x 10^-4 C

b) u= 1/2 cv^2=0.5 ( 4 x 10^-6) ( 100^2) = 2 x 10^ -2 J

c) charge will flow to second capacitors, till they both acquire equal potential

Q1/C1= Q2/C2

Therfore charge on second capacitor will be 2 x 10^ -4 C

so that total charge remains 4 x10^-4 C

d) Qt= Ct V

4 x10^-4 = 8 x 10^-6 x V

v = 50 V

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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