Question 154: Electrostats & Capacitance

Question

A parallel plate capacitor of capacitance $$2 \mathrm{~F}$$ is charged to a potential $$\mathrm{V}$$, The energy stored in the capacitor is $$E_{1}$$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $$\mathrm{E}_{2}$$. The ratio $$\mathrm{E}_{2} / \mathrm{E}_{1}$$ is :

Accepted Answer

To determine the ratio $$\mathrm{E}_{2} / \mathrm{E}_{1}$$, we will follow these steps:
1. Calculate the initial energy stored in the original capacitor $$\mathrm{E}_{1}$$.
2. Determine the energy stored in the system when two capacitors are connected in parallel, which is $$\mathrm{E}_{2}$$.
3. Find the ratio $$\mathrm{E}_{2} / \mathrm{E}_{1}$$.
Step 1: Calculate the initial energy stored in the original capacitor $$\mathrm{E}_{1}$$.
The energy stored in a capacitor is given by the formula:
$$E = \frac{1}{2} CV^2$$
Given that the capacitance $$C$$ is $$2 \mathrm{~F}$$, and the potential difference is $$\mathrm{V}$$, the initial energy $$\mathrm{E}_{1}$$ is:
$$E_{1} = \frac{1}{2} \cdot 2 \mathrm{~F} \cdot V^2$$
$$E_{1} = V^2 \mathrm{~J}$$
Step 2: Determine the energy stored when two capacitors are connected in parallel
When the charged capacitor (capacitor 1) is connected to an identical uncharged capacitor (capacitor 2), the charge will redistribute between the two capacitors. The total capacitance of the parallel combination is:
$$C_{	ext{total}} = 2 \mathrm{~F} + 2 \mathrm{~F} = 4 \mathrm{~F}$$
The initial charge on capacitor 1 is:
$$Q_1 = CV = 2 \mathrm{~F} \cdot V = 2V \mathrm{~C}$$
After connection, this charge will be shared equally by the two capacitors because they are identical. Therefore, the voltage across each capacitor in the parallel combination will be:
$$V_{	ext{across each capacitor}} = \frac{	ext{Total charge}}{	ext{Total capacitance}} = \frac{2V}{4 \mathrm{~F}} = \frac{V}{2}$$
The energy stored in the parallel combination is:
$$E_{2} = \frac{1}{2} \cdot 4 \mathrm{~F} \cdot \left(\frac{V}{2}ight)^2$$
$$E_{2} = \frac{1}{2} \cdot 4 \mathrm{~F} \cdot \frac{V^2}{4}$$
$$E_{2} = V^2 \cdot \frac{1}{2} \mathrm{~J}$$
Step 3: Find the ratio $$\mathrm{E}_{2} / \mathrm{E}_{1}$$
We know that:
$$E_{1} = V^2 \mathrm{~J}$$
$$E_{2} = \frac{1}{2} V^2 \mathrm{~J}$$
The ratio is then:
$$\frac{E_{2}}{E_{1}} = \frac{\frac{1}{2} V^2}{V^2} = \frac{1}{2} = 1 : 2$$
Therefore, the correct answer is:
Option A: 1 : 2

Answer

1 : 2

Answer

2 : 3

Answer

2 : 1

Answer

1 : 4

Electrostats & Capacitance

Choose the correct answer: