Physics

Problem

Capacitors

C_1=10\,\mu\text{F}

and

C_2=20\,\mu\text{F}

are connected in parallel. This parallel combination is connected in series with a capacitor

C_3=5\,\mu\text{F}

across a

30\,\text{V}

ideal battery. After the capacitors are fully charged, the battery is disconnected from the circuit without discharging the capacitors. A dielectric slab of dielectric constant

\kappa = 2

is then completely inserted between the plates of

C_3

, fully filling the gap. a. (i) Calculate the charge on each capacitor immediately before the battery is disconnected. b. (ii) Determine the new value of the capacitance of

C_3

after the dielectric is inserted. c. (iii) Find the new potential difference across

C_3

and the common potential difference across the parallel branch containing

C_1

and

C_2

." d. (iv) Compute the total energy stored in the system (a) before and (b) after the insertion of the dielectric. Hence determine the change in energy of the system.

Intuitive explanation

When capacitors are in parallel, they act as a single larger capacitor; when that combination sits in series with another, the same charge flows through both branches. The key insight is that after disconnecting the battery, the total charge in the circuit is frozen—it cannot change. Inserting the dielectric increases $C_3$ 's capacitance, which redistributes voltage between the series branches while keeping total charge constant. Use $Q = CV$ and charge conservation to find the new voltage across each section and the energy released.

Intuitive explanation

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