Two capacitors A and B with capacities 3 uF and 2 uF are charged to a potential difference of 100 V

Q: Why do we need to worry about the polarities of the initial charges?

The polarities determine whether the charges on connected plates will add up or partially cancel each other out. If you connect a positive plate to a negative plate, the net charge is their algebraic sum, which drastically changes the final equilibrium state.

Q: How do we identify an isolated system in a capacitor circuit?

An isolated system consists of plates and connecting wires that are completely cut off from the rest of the circuit by the dielectric gaps of the capacitors. No charge can jump across the dielectric, so the total charge trapped in this section must remain constant.

Q: Why is the potential drop across capacitor B written as $+q_3/C_B$ in the KVL equation?

When traversing the loop clockwise, we go from the top plate of B (which we defined as having $-q_3$ charge) to the bottom plate (which has $+q_3$ charge). Moving from a negative to a positive plate means we are moving to a higher potential, hence the positive sign.

Q: Why is energy lost when the circuit is completed?

When the charges redistribute, they flow through the connecting wires. Even if the wires have negligible resistance, the sudden movement of charge creates a transient current that dissipates energy as heat and electromagnetic radiation. This is a classic characteristic of capacitor redistribution problems.

Q: Could we have solved this using the concept of common potential?

The standard common potential formula $V = (C_1V_1 + C_2V_2)/(C_1 + C_2)$ only works for two capacitors connected in parallel. Here, we have three capacitors forming a loop, so we must rely on the fundamental principles of charge conservation and KVL.

Question

Two capacitors $A$ and $B$ with capacities $3\,\mu	ext{F}$ and $2\,\mu	ext{F}$ are charged to a potential difference of $100	ext{ V}$ and $180	ext{ V}$ respectively. The plates of the capacitors are connected as shown in the figure with one wire of each capacitor free. The upper plate of $A$ is positive and that of $B$ is negative. An uncharged $2\,\mu	ext{F}$ capacitor $C$ with lead wires falls on the free ends to complete the circuit. Calculate
(a) the final charge on the three capacitors and
(b) the amount of electrostatic energy stored in the system before and after completion of the circuit.

Sigma Percentile · Accepted Answer

The problem of redistributing charges among multiple capacitors is a classic test of your understanding of fundamental electrostatic principles. It's not just about plugging numbers into a formula; it's about tracing the journey of every single electron. Let's dive into this beautiful circuit!

### Analyzing the Setup

Before the uncharged capacitor $C$ falls and completes the circuit, capacitors $A$ and $B$ are sitting there, fully charged and isolated. Our first task is to figure out exactly how much charge they are holding.

Using the fundamental relation $Q = CV$, we can calculate the initial charges:
For capacitor $A$:
$$q_A = C_A V_A = (3\,\mu	ext{F})(100	ext{ V}) = 300\,\mu	ext{C}$$

For capacitor $B$:
$$q_B = C_B V_B = (2\,\mu	ext{F})(180	ext{ V}) = 360\,\mu	ext{C}$$

**Crucial Step:** We must pay close attention to the polarities! The problem states the upper plate of $A$ is positive ($+300\,\mu	ext{C}$), and the upper plate of $B$ is negative ($-360\,\mu	ext{C}$). This asymmetry is the trap where many students stumble.

### The Master Equations: Charge Conservation

When capacitor $C$ drops in, the circuit is completed. Charges will rush through the wires, seeking a new equilibrium. Let's call the final charges on the three capacitors $q_1$, $q_2$, and $q_3$ respectively.

How do we find them? We look for **isolated systems**. An isolated system is a section of the circuit bounded entirely by the dielectric gaps of the capacitors. No charge can cross a dielectric, so the total charge trapped in that section must remain constant forever.

**Isolated System 1:** Look at the top plate of $A$ and the left plate of $C$. They are connected by a wire, but isolated from everything else.
Initial charge = $+300\,\mu	ext{C}$ (from $A$) + $0$ (from $C$) = $+300\,\mu	ext{C}$.
Final charge = $+q_1$ (on $A$) + $+q_2$ (on $C$).
$$q_1 + q_2 = 300$$

**Isolated System 2:** Now look at the right plate of $C$ and the top plate of $B$.
Initial charge = $0$ (from $C$) + $-360\,\mu	ext{C}$ (from $B$) = $-360\,\mu	ext{C}$.
Final charge = $-q_2$ (on $C$) + $-q_3$ (on $B$).
$$-q_2 - q_3 = -360 \implies q_2 + q_3 = 360$$

### The Loop Rule: Kirchhoff's Voltage Law

We have three variables but only two equations. We need one more constraint: the fact that the total potential drop around any closed loop must be zero. Let's apply Kirchhoff's Voltage Law (KVL) clockwise around the loop.

Starting from the bottom wire and going up through $A$, we move from the negative to the positive plate. The potential increases by $q_1/C_A$.
Moving right through $C$, we go from positive to negative. The potential drops by $q_2/C_C$.
Moving down through $B$, we go from the negative top plate to the positive bottom plate. The potential increases by $q_3/C_B$.

Summing these up:
$$\frac{q_1}{3} - \frac{q_2}{2} + \frac{q_3}{2} = 0$$

Multiplying the entire equation by 6 to clear the denominators:
$$2q_1 - 3q_2 + 3q_3 = 0$$

### Final Calculation of Charges

Now we have a beautiful system of linear equations. Let's express $q_1$ and $q_3$ in terms of $q_2$ using our conservation equations:
$q_1 = 300 - q_2$
$q_3 = 360 - q_2$

Substitute these into the KVL equation:
$$2(300 - q_2) - 3q_2 + 3(360 - q_2) = 0$$
$$600 - 2q_2 - 3q_2 + 1080 - 3q_2 = 0$$
$$1680 - 8q_2 = 0$$
$$8q_2 = 1680 \implies q_2 = 210\,\mu	ext{C}$$

With $q_2$ unlocked, the rest is trivial:
$$q_1 = 300 - 210 = 90\,\mu	ext{C}$$
$$q_3 = 360 - 210 = 150\,\mu	ext{C}$$

### The Energy Audit

The second part of the question asks for the energy before and after. 
The initial energy is simply the sum of the energies of the isolated capacitors:
$$U_i = \frac{1}{2} C_A V_A^2 + \frac{1}{2} C_B V_B^2$$
$$U_i = \frac{1}{2} (3 	imes 10^{-6}) (100)^2 + \frac{1}{2} (2 	imes 10^{-6}) (180)^2$$
$$U_i = 1.5 	imes 10^{-2} + 3.24 	imes 10^{-2} = 4.74 	imes 10^{-2}	ext{ J} = 47.4	ext{ mJ}$$

The final energy is calculated using the new charges we just found:
$$U_f = \frac{q_1^2}{2C_A} + \frac{q_2^2}{2C_C} + \frac{q_3^2}{2C_B}$$
$$U_f = \frac{(90 	imes 10^{-6})^2}{2(3 	imes 10^{-6})} + \frac{(210 	imes 10^{-6})^2}{2(2 	imes 10^{-6})} + \frac{(150 	imes 10^{-6})^2}{2(2 	imes 10^{-6})}$$
$$U_f = (1350 + 11025 + 5625) 	imes 10^{-6}	ext{ J} = 18000 	imes 10^{-6}	ext{ J} = 18	ext{ mJ}$$

Notice the massive drop in energy! We started with $47.4	ext{ mJ}$ and ended with only $18	ext{ mJ}$. A total of $29.4	ext{ mJ}$ was dissipated. This energy is lost as heat in the connecting wires and radiated away as electromagnetic waves during the sudden charge redistribution. It's a stark reminder that while charge is always conserved, mechanical/electrical energy in dynamic processes often is not!

Visualized Solution

$I ni t ia l C ha r g es$

$C ha r g e R e d i s t r ib u t i o n$

$C o n ser v a t i o n o f C ha r g e$

$K i r c hh o f f^{'} s V o l t a g e L a w$

$S o l v in g f or q_{2}$

$F ina l C ha r g es$

$I ni t ia l E n er g y$

$F ina l E n er g y$

$T h e W a y F or w a r d$

The Sigma Insight: Combination of Capacitors

Analyzing the Setup

The Master Equations

The Loop Rule

Final Calculation of Charges

The Energy Audit

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Charges

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Charge Redistribution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Charge

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kirchhoff′s Voltage Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for q2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Charges

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward