A 3.55 mu F capacitor, C_1, is charged to a potential difference V_0 = 6.30 C, u

Question

A 3.55 mu F capacitor, C_1, is charged to a potential difference V_0 = 6.30 C, using a battery. The charging battery is then removed, and the capacitor is connected (as in the figure to the right) to an uncharged 8.95 mu F capacitor C_3. After the switch, is closed, charge flows from C_1 to C_2 until an equilibrium is established, with both capacitors at the same potential difference, What is this common potential difference? What is the energy stored in the electric field before and after the switch in the figure is thrown?

Explanation / Answer

a)

Common potential , V = Total charge / Total capacitance = Vo C1 / (C1 + C2) = (6.30) (3.55)/(3.55 + 8.95)

V = 1.789

b)

before switch closed

E = (0.5) C1 Vo2 = (0.5) (3.55 x 10-6) (6.30)2 = 0.0000704 J

after switch closed

E = (0.5) (C1 + C2) V2 = (0.5) (3.55 x 10-6 + 8.95 x 10-6) (1.789)2 = 0.00002 J

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