Two parallel-plate capacitors, each having a capacitance of 2.0 mu F are connect

Question

Two parallel-plate capacitors, each having a capacitance of 2.0 mu F are connected in parallel across a 12-V battery. Find the charge on each capacitor. Find the total energy stored in the capacitors. The parallel combination is then disconnected from the battery and a dielectric slab of constant 2.5 is inserted between the plates of the second capacitor (C_2), completely filling the gap. After the dielectric is inserted, find the potential difference across each capacitor, the charge on each capacitor, and the total energy stored in the capacitors.

Explanation / Answer

in parallel total capacitance

C = C1+ C2 = 2 + 2 = 4uF

Q = CV

Q = 2uF * 12 = 24 uF

part b )

U = 1/2*Ceq*V^2

U = 1/2*4*10^-6 *12^2

U = 2.9 x 10^-4 J

part c )

now total capacitance = 2 + 5 = 7uF

C2 = k*C = 2.5 *2 = 5 uF

Q remain same

Q toal = 48 uC

V = Q/C = 6.86 V

in parallel voltage remain same

part d )

Q1 = C1V

Q1 = 2 * 10^-6 x 6.86 = 13.72 * 10^-6 = 13.72 uC

Q2 = C2V

Q2 = 5 * 10^-6 x 6.86 = 34.3 * 10^-6 = 34.3 uC

part e )

U = 1/2*CV^2

U = 1/2*7*10^-6 x 6.86^2

U = 1.65 x 10^-4 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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