A capacitor C_1 = 4.00 mu F is charged up to Delta V = 100 Volt. Another capacit

Question

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). a) What is the chare q on C_1 before both capacitors are connected? b) What is the energy U stored on C_1 before both capacitors are connected? c) 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) For a charge q given to the assembly

q=C1V
where V is voltage drop across capacitors

q=4x10-6 x 100

=4x10-4Coulombs

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

b)The energy stored on a capacitor can be calculated from the equivalent expressions:

U=1/2 Cv2  This energy is stored in the electric field.

U=1/2 x 4x10-6 x 1002

= .02 Joules

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

c)Capacitors in parallel have the same voltage. Charge may redistribute among them: Q1/C1 = Q2/C2.......

already find q1=4x10-4Coulombs

since its parallel voltage is same that is 100v

q2=C2v= 4x10-6 x 100= 4x10-4Coulombs

Total charge q'=q1+q2=2x 4x10-4 = 8x10-4Coulombs

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

d)New voltage will be the same 100v since its a parallel connection.voltage will be same

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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