Question

A variable condenser is permanently connected to a \(100 \mathrm{~V}\) battery. If capacitor is changed from \(2 \mu \mathrm{F}\) to \(10 \mu \mathrm{F}\). then energy changes is equal to (A) \(2 \times 10^{-2} \mathrm{~J}\) (B) \(2.5 \times 10^{-2} \mathrm{~J}\) (C) \(6.5 \times 10^{-2} \mathrm{~J}\) (D) \(4 \times 10^{-2} \mathrm{~J}\)

Short answer

The change in energy when the capacitor is changed from $2 \mu \mathrm{F}$ to $10 \mu \mathrm{F}$ is \(\boldsymbol{4 \times 10^{-2} \mathrm{~J}}\).

Step-by-step solution

Calculate the energy for the first capacitor

The energy stored in a capacitor can be calculated using the following formula: \[E = \frac{1}{2}CV^2\] where C is the capacitance, and V is the voltage across the capacitor. For the first capacitor with a capacitance of 2 μF, the energy stored can be calculated as: \(E_1 = \frac{1}{2}(2\times10^{-6}\,\mathrm{F})(100\,\mathrm{V})^2\)

Calculate the energy for the second capacitor

Similarly, for the second capacitor with a capacitance of 10 μF, the energy stored can be calculated using the same formula: \(E_2 = \frac{1}{2}(10\times10^{-6}\,\mathrm{F})(100\,\mathrm{V})^2\)

Calculate the change in energy

Now, we will calculate the change in energy by finding the difference between the two energy values: ΔE = E₂ - E₁

Substitute the values and find the answer

Substitute the values of E₁ and E₂ in the formula, and calculate the change in energy: ΔE = \(\frac{1}{2}(10\times10^{-6}\,\mathrm{F})(100\,\mathrm{V})^2\) - \(\frac{1}{2}(2\times10^{-6}\,\mathrm{F})(100\,\mathrm{V})^2\) ΔE = \(4\times10^{-2}\,\mathrm{J}\) Hence, the change in energy is \(\boldsymbol{4 \times 10^{-2} \mathrm{~J}}\), which corresponds to option (D).

Key concepts

Capacitance Change

When we talk about capacitance change, we're looking at the variation in the ability of a capacitor to store charge as its capacitance value alters. Capacitance ( C ) is quantified in farads ( F ) and indicates how much electric charge a capacitor can store for a given voltage.
In our exercise, the capacitor's capacitance is changed from 2 microfarads ( μF ) to 10 microfarads. This change significantly impacts the amount of energy the capacitor can store.
The higher the capacitance, the more charge the capacitor can store for the same voltage, leading to a higher energy storage.

• Initial Capacitance: 2 μF
• Final Capacitance: 10 μF

This increase in capacitance will lead to a change in the energy stored, illustrating why understanding capacitor properties is key to comprehending changes in energy dynamics within electronic circuits.

Voltage Across Capacitor

The voltage across a capacitor plays an essential role in determining how much energy it can store. This voltage is basically the potential difference between the plates of the capacitor.
In our scenario, the voltage remains constant at 100 volts (V), even when the capacitance changes. By keeping the voltage constant, we can examine the sole effect of the change in capacitance on the energy stored.

• Constant Voltage: 100 V

Even though the capacitance increases, the voltage does not, which simplifies our calculations and helps us focus on the capacitance effects. A consistent voltage ensures that the changes we observe in energy storage are due to capacitance variations.

Energy Calculation

Calculating energy stored in a capacitor involves the equation \[E = \frac{1}{2}C V^2\]where E is the energy, C is the capacitance, and V is the voltage.
In our physics problem, we first calculate the energy stored for each capacitance value separately.

• Energy with 2 μF: \[E_1 = \frac{1}{2}(2\times10^{-6}\,\mathrm{F})(100\,\mathrm{V})^2\]
• Energy with 10 μF: \[E_2 = \frac{1}{2}(10\times10^{-6}\,\mathrm{F})(100\,\mathrm{V})^2\]

Then, the difference, ΔE, represents the change in energy due to the change in capacitance.
Once calculated, this difference highlights how much more energy the capacitor can store with the increased capacitance, assuming other factors remain constant, a concept that is frequently encountered in electrical engineering and physics.

Physics Problem Solving

Solving physics problems effectively requires a methodical approach. Let's break down the steps:
First, understand the problem and identify what is being asked. In this case, it's the change in energy due to capacitance change. Next, identify and write down all given information, like initial and final capacitance and the constant voltage.
Decide on the relevant formulas. For capacitor energy, the formula \[E = \frac{1}{2}C V^2\] is used.
Calculate intermediate values, such as energy for each capacitance, before moving to the next part of the problem. Finally, evaluate the change in energy. Each of these steps systematically builds upon the last, leading you to the solution efficiently. Solving physics problems isn't just about getting the right answer, but understanding the process and relationships between variables, which deepens your overall comprehension of how concepts like energy and capacitance interact.