Question 148: Electrostats & Capacitance

Question

Two identical capacitors have same capacitance $C$. One of them is charged to the potential $V$ and other to the potential $2 \mathrm{~V}$. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :

Accepted Answer

To solve the problem, let's start by considering the energy stored in each capacitor before they are connected together.
The energy stored in a capacitor is given by the formula:
$$E = \frac{1}{2} C V^2$$
Where $E$ is the energy, $C$ is the capacitance, and $V$ is the potential.
For the first capacitor charged to the potential $V$, the energy stored is:
$$E_1 = \frac{1}{2} C V^2$$
For the second capacitor charged to the potential $2V$, the energy stored is:
$$E_2 = \frac{1}{2} C (2V)^2 = \frac{1}{2} C \cdot 4V^2 = 2 CV^2$$
The total initial energy stored in the system is the sum of $E_1$ and $E_2$:
$$E_{	ext{initial}} = E_1 + E_2 = \frac{1}{2} CV^2 + 2 CV^2 = \frac{5}{2} CV^2$$
When the two capacitors are connected together, their potentials will become equal because they are identical capacitors. Let's denote this final potential as $V_f$. The total charge before and after the connection remains constant because charge is conserved. Therefore, we can write:
$$C \cdot V + C \cdot 2V = 2C \cdot V_f$$
Simplifying this equation gives us the final potential:
$$V + 2V = 2 V_f$$
$$3V = 2 V_f$$
$$V_f = \frac{3}{2} V$$
The final energy stored in the system when the capacitors are connected is now the total energy stored across both capacitors at the final potential $V_f$:
$$E_{	ext{final}} = 2 \cdot \frac{1}{2} C V_f^2 = 2 \cdot \frac{1}{2} C \left(\frac{3}{2} Vight)^2 = C \cdot \frac{9}{4} V^2$$
The decrease in energy $ \Delta E $ of the combined system is the initial energy minus the final energy:
$$\Delta E = E_{	ext{initial}} - E_{	ext{final}}$$
$$\Delta E = \frac{5}{2} CV^2 - \frac{9}{4} CV^2$$
$$\Delta E = \frac{10}{4} CV^2 - \frac{9}{4} CV^2$$
$$\Delta E = \frac{1}{4} CV^2$$
The correct answer is:
Option A: $\frac{1}{4} CV^2$

Answer

$\frac{1}{4} \mathrm{CV}^2$

Answer

$\frac{3}{4} \mathrm{CV}^2$

Answer

$\frac{1}{2} \mathrm{CV}^2$

Answer

$2 \mathrm{CV}^2$

Electrostats & Capacitance

Two identical capacitors have same capacitance C. One of them is charged to the potential V and other to the potential 2 \mathrm{~V}. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :

Choose the correct answer: