A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance w

Q: Why doesn't the resistance of the coil appear in the final formula?

The resistance determines *how fast* the capacitor discharges, but not the *total amount* of energy dissipated. By the time it's fully discharged, all stored energy has become heat, regardless of the resistance.

Q: What if the block wasn't thermally insulated?

If it wasn't insulated, some heat would escape into the surroundings. The temperature rise $\Delta T$ would be smaller, and our simple energy equation wouldn't hold without accounting for the lost heat.

Q: Why do we use $E = \frac{1}{2}CV^2$ instead of $E = \frac{Q^2}{2C}$?

Both formulas are perfectly valid! However, since the question specifically asks for the potential difference $V$, using the formula that already contains $V$ saves us an extra substitution step.

Q: Does the specific heat capacity $s$ depend on the temperature?

For small temperature changes, we assume $s$ is constant. In reality, it can vary slightly with temperature, but in standard physics problems, we treat it as a fixed property of the material.

Q: Where does the energy actually come from?

The energy comes from the electric field established between the plates of the charged capacitor. As electrons flow through the circuit to neutralize the charge, this field collapses, and its energy is transferred to the electrons, which then collide with atoms in the coil, generating heat.

Question

A fully charged capacitor has a capacitance $C$. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity $s$ and mass $m$. If the temperature of the block is raised by $\Delta T$, the potential difference $V$ across the capacitance is

Options:

(A) $\sqrt{\frac{2mC\Delta T}{s}}$

(B) $\frac{mC\Delta T}{s}$

(C) $\frac{ms\Delta T}{C}$

(D) $\sqrt{\frac{2ms\Delta T}{C}}$

Sigma Percentile · Accepted Answer

### The Setup: A Trapped Spark

Imagine a capacitor that is fully charged, brimming with electrostatic potential energy. It is like a water tank filled to the brim, just waiting for a valve to open. When we connect this capacitor to a resistance coil, the "valve" opens, and it starts to discharge. The electrons rush through the circuit, creating a current.

But here is the twist: this resistance coil is embedded inside a thermally insulated block. This means whatever happens inside the block, stays inside the block. No energy can escape into the surrounding environment.

### The Master Equation: Conservation of Energy

How much energy does the capacitor have initially? The energy stored in a charged capacitor is given by the elegant formula:

$$E = \frac{1}{2}CV^2$$

This is the total energy that will be released during the discharge process. Now, as the current forces its way through the resistance coil, it encounters opposition. This electrical friction generates heat. This heat will gradually raise the temperature of the block.

From thermodynamics, we know that the heat $H$ required to raise the temperature of a block of mass $m$ and specific heat capacity $s$ by an amount $\Delta T$ is:

$$H = ms\Delta T$$

Since the block is perfectly thermally insulated, the universe demands that energy be conserved locally. The entire electrical energy of the capacitor is completely converted into the thermal energy of the block. Therefore, we can confidently equate the two expressions:

$$\frac{1}{2}CV^2 = ms\Delta T$$

### The Final Calculation: Unveiling the Voltage

Our goal is to find the initial potential difference, $V$, across the capacitor. Let's isolate $V$ by rearranging our conservation equation. First, we multiply both sides by 2 and divide by $C$:

$$V^2 = \frac{2ms\Delta T}{C}$$

Finally, taking the square root of both sides reveals the answer:

$$V = \sqrt{\frac{2ms\Delta T}{C}}$$

**This beautiful result bridges two entirely different realms of physics: electrostatics and thermodynamics.** It shows exactly how the invisible electric field translates into the physical vibration of atoms, raising the temperature of the block.

Select Answer:

Visualized Solution

Visualizing the Setup

Initial Electrical Energy

Heat Generated in the Block

Conservation of Energy

Solving for Voltage

Final Answer

The Way Forward

The Sigma Insight: Capacitance and Capacitors

The Setup

The Master Equation

The Final Calculation

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