Question

Two identical capacitors have the same capacitance \(\mathrm{C}\). one of them is charged to a potential \(\mathrm{V}_{1}\) and the other to \(\mathrm{V}_{2}\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is (A) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}\right)\) (B) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}+\mathrm{V}_{2}^{2}\right)\) (C) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}\) (D) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)^{2}\)

Short answer

The short answer to the given problem is: The decrease in energy of the combined system is \( \frac{1}{4}C(V_1 - V_2)^2 \). The correct option is (C).

Step-by-step solution

Find initial energies of the capacitors

To find the initial energy of each capacitor, we can use the energy formula for capacitors: \[E = \frac{1}{2} CV^2\] The initial energy of the first capacitor is: \[E_1 = \frac{1}{2} CV_1^2\] The initial energy of the second capacitor is: \[E_2 = \frac{1}{2} CV_2^2\] The total initial energy of the system is: \[E_{initial} = E_1 + E_2 = \frac{1}{2}C\left(V_1^2 + V_2^2\right)\]

Find the final energy of the combined system

When the capacitors are connected, they will have the same voltage V across them. Since the capacitors are identical and have the same capacitance C, the total capacitance of the connected capacitors remains C. The charge conservation principle states that the sum of initial charges equals the sum of final charges. Therefore: \[Q_1 + Q_2 = Q_f\] Using the formula for the charge of a capacitor (Q = CV), we get: \[CV_1 + CV_2 = 2CV_f\] \[V_f = \frac{V_1 + V_2}{2}\] The final energy of the combined system is: \[E_{final} = \frac{1}{2}(2C)\left( \frac{V_1 + V_2}{2}\right)^2 = \frac{1}{4}C\left(V_1 + V_2\right)^2\]

Calculate the decrease in energy and find the correct answer

The decrease in energy of the combined system is: \[∆E = E_{initial} - E_{final} = \frac{1}{2}C\left( V_1^2 + V_2^2\right) - \frac{1}{4}C\left(V_1 + V_2\right)^2\] \[= \frac{1}{4}C\left[2(V_1^2 + V_2^2) - (V_1 + V_2)^2\right]\] \[= \frac{1}{4}C\left[2(V_1^2 + V_2^2) - (V_1^2 + 2V_1V_2 + V_2^2)\right]\] \[= \frac{1}{4}C\left(V_1^2 - 2V_1V_2 + V_2^2\right)\] \[= \frac{1}{4}C\left(V_1 - V_2\right)^2\] The decrease in energy of the combined system is given by option C, so the correct answer is C: \((1 / 4) C (V_1 - V_2)^2\).

Key concepts

Capacitance

Capacitance is a fundamental concept in physics and electrical engineering. It refers to a capacitor's ability to store an electric charge. A capacitor consists of two conductive plates separated by an insulating material known as a dielectric. When a voltage is applied across these plates, an electric field develops, causing an accumulation of electric charge on the plates. The capacitance of a capacitor, denoted by the letter \(C\), is expressed in farads and is calculated using the formula:- \(C = \frac{Q}{V}\) - \(Q\) is the charge stored on the plates. - \(V\) is the voltage across the plates.Higher capacitance values mean the capacitor can store more charge at a given voltage. Capacitance is determined by factors such as the surface area of the plates, the distance between them, and the properties of the dielectric material.

Energy in Capacitors

Understanding how capacitors store energy is crucial for many electronic applications. The energy \(E\) stored in a capacitor is given by the formula:- \(E = \frac{1}{2} CV^2\) - \(C\) is the capacitance. - \(V\) is the voltage across the capacitor.When the voltage across a capacitor changes, the energy stored also changes. In a circuit, this energy can be released to power a device or to perform work, making capacitors useful for stabilizing voltage levels and providing power bursts in electronic systems. When two capacitors with different voltages are connected, they redistribute their charges resulting in a new voltage and a change in stored energy. The loss or decrease in energy is part of how capacitors balance out the system when connected together.

Charge Conservation

Charge conservation is a fundamental principle often applied when dealing with capacitors in circuits. It states that the total electric charge in an isolated system remains constant over time. When two capacitors are joined, the charge on them redistributes to reach equilibrium.Mathematically, if you have two capacitors with charges \(Q_1\) and \(Q_2\) initially, the combined charge \(Q_f\) after connecting them is:- \(Q_f = Q_1 + Q_2\)This means no charge is lost in the system; it's merely rearranged between the components. The principle helps in calculating the final voltage or energy state when capacitors are combined, utilizing formulas like:- \(CV_1 + CV_2 = 2CV_f\) - Where \(V_f\) represents the final voltage after combining the capacitors.

Voltage in Circuits

Voltage is a crucial element in understanding how capacitors and circuits operate. It is the potential difference between two points in an electric field or circuit, essentially measuring the energy per charge unit. Voltages drive the movement of charge around the circuit and can be thought of as the 'pressure' pushing the charge.When capacitors are involved, the voltage across each capacitor affects how charges are stored and redistributed. In a closed loop of capacitors, the voltage tends to equalize across all connected components due to charge redistribution.In the exercise above, when capacitors are connected, they share their voltages, leading to an average voltage calculation. Calculating this average, indicated as \(\overline{V}\), is crucial:- \(\overline{V} = \frac{V_1 + V_2}{2}\) This average dictates the new voltage across each capacitor once the connection is made, affecting their charge and energy in turn.