A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is

Q: Why is energy lost if charge is conserved?

Charge is conserved, but energy is not. When charge flows through the connecting wires, it encounters resistance. This resistance causes some of the electrical energy to be dissipated as heat (Joule heating). Even if the wires were perfectly superconducting, the energy would be radiated away as electromagnetic waves!

Q: Does the amount of energy lost depend on the resistance of the wires?

Surprisingly, no! The *rate* at which the energy is lost depends on the resistance, but the *total amount* of energy lost is completely independent of the resistance. It only depends on the initial and final states of the capacitors.

Q: Why did the potential drop to exactly half?

Because the two capacitors have the exact same capacitance ($60\text{ pF}$). When they are connected in parallel, the total charge has to spread out over twice the original capacitance. Since $V = Q/C$, doubling the capacitance while keeping the charge constant halves the potential.

Q: Can I use a direct formula to find the energy loss?

Yes! The direct formula is $\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$. It's a massive time-saver for competitive exams like JEE and NEET.

Q: What if the second capacitor was also charged initially?

The same principles apply. You would find the total initial charge ($Q_1 + Q_2$), calculate the common potential, and then find the difference between the total initial energy and the total final energy. The direct formula mentioned above also works perfectly in this case!

Question

A $60 	ext{ pF}$ capacitor is fully charged by a $20 	ext{ V}$ supply. It is then disconnected from the supply and is connected to another uncharged $60 	ext{ pF}$ capacitor in parallel. The electrostatic energy that is lost in this process by the time, the charge is redistributed between them is (in $	ext{nJ}$)

Sigma Percentile · Accepted Answer

### The Mystery of the Missing Energy: Capacitor Charge Redistribution

Have you ever wondered what happens when you connect a fully charged capacitor to an uncharged one? It seems like a simple transfer of charge, but there is a hidden mystery: some energy always goes missing! Let's dive into this classic electrostatics problem and uncover where that energy goes.

### The Initial State

Imagine our initial setup. We have a capacitor with a capacitance of $C_1 = 60	ext{ pF}$ connected to a $V_1 = 20	ext{ V}$ battery. It acts like a tiny electrical storage tank, filling up with charge until it reaches the battery's voltage.

The energy stored in this fully charged capacitor can be calculated using the standard formula:

$$U_1 = \frac{1}{2} C_1 V_1^2$$

Let's substitute the values. Remember to convert picofarads to farads by multiplying by $10^{-12}$:

$$U_1 = \frac{1}{2} 	imes (60 	imes 10^{-12}) 	imes (20)^2$$

$$U_1 = 12 	imes 10^{-9}	ext{ J} = 12	ext{ nJ}$$

So, our initial energy bank holds exactly $12	ext{ nJ}$.

### The Connection and Redistribution

Now, the battery is removed, and this charged capacitor is connected in parallel to another identical, but uncharged capacitor ($C_2 = 60	ext{ pF}$). Because they are in parallel, charge will flow from the charged capacitor to the uncharged one. This flow continues until they both reach the exact same electrical pressure, known as the **common potential ($V$)**.

This common potential is simply the total charge divided by the total capacitance:

$$V = \frac{Q_{total}}{C_{total}} = \frac{C_1 V_1 + 0}{C_1 + C_2}$$

Since both capacitors are $60	ext{ pF}$, the total charge has to spread out over twice the original capacitance. Consequently, the common potential becomes exactly half of the initial voltage:

$$V = \frac{60 	imes 20}{60 + 60} = 10	ext{ V}$$

### The Final State and The Missing Energy

Now, let's find the final energy of this combined system. The equivalent capacitance is the sum of the two ($120	ext{ pF}$), and we use our newly found common potential of $10	ext{ V}$.

$$U_2 = \frac{1}{2} C_{eq} V^2$$

$$U_2 = \frac{1}{2} 	imes (120 	imes 10^{-12}) 	imes (10)^2$$

$$U_2 = 60 	imes 10^{-12} 	imes 100 = 6 	imes 10^{-9}	ext{ J} = 6	ext{ nJ}$$

Wait a minute! We started with $12	ext{ nJ}$, and now we only have $6	ext{ nJ}$. Where did the other $6	ext{ nJ}$ go?

$$\Delta U = U_1 - U_2 = 12	ext{ nJ} - 6	ext{ nJ} = 6	ext{ nJ}$$

This energy is **lost as heat** in the connecting wires due to their electrical resistance. Even if the wires were perfectly superconducting, the energy would still be lost, radiated away as electromagnetic waves during the sudden surge of charge!

### The Master Formula

As a pro tip for competitive exams, you can bypass these steps and use a direct formula for energy loss when two capacitors are connected:

$$\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$$

Try plugging in our values ($V_2 = 0$ since the second capacitor was uncharged). You will get the exact same result of $6	ext{ nJ}$ in a fraction of the time!

Enter Numerical Value:

Visualized Solution

$C_{1} = 60 pF, V_{1} = 20 V$

$U_{1} = \frac{1}{2} C_{1} V_{1}^{2}$

$U_{1} = 12 nJ$

$C_{2} = 60 pF$

$V = \frac{Q _{t o t a l}}{C _{t o t a l}}$

$V = 10 V$

$U_{2} = \frac{1}{2} C_{e q} V^{2}$

$U_{2} = 6 nJ$

$Δ U = U_{1} - U_{2}$

$Δ U = \frac{1}{2} \frac{C _{1} C _{2}}{C _{1} + C _{2}} (V_{1} - V_{2})^{2}$

The Sigma Insight: Combination of Capacitors

The Mystery of the Missing Energy

The Initial State

The Connection and Redistribution

The Final State and The Missing Energy

The Master Formula

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​=60 pF,V1​=20 V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U1​=21​C1​V12​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U1​=12 nJ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​=60 pF

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=Ctotal​Qtotal​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=10 V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U2​=21​Ceq​V2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U2​=6 nJ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔU=U1​−U2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔU=21​C1​+C2​C1​C2​​(V1​−V2​)2