Imagine a camera flash. If you press the button, it produces light. The flash requires additional energy to produce that light. This energy is stored in a Capacitor inside the camera. the capacitor charges up when it is connected to the battery and energy is stored in it and then gives out all energy, thus creating the flash. In this article, we will study Capacitor and its working. We will also study the expression of energy stored in capacitors. In the later part, we will derive the equation of energy stored in a capacitor followed by work done by the capacitor.
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A capacitor is a device consisting of two conductive plates separated by an insulating material known as a Dielectric. Basically, it is an electric component that stores charge in the form of an electric field.
Capacitance is measured in farads (f).
1 farad ( $F$ ) is defined as the capacitance of a capacitor that, when charged with 1 coulomb ( $C$ ) of electric charge, results in a potential difference of 1 volt $(\mathrm{V})$ across its plates:
$$
1 \mathrm{~F}=\frac{1 \mathrm{C}}{1 \mathrm{~V}}
$$
Thus capacitance is given by the formula:
$$
\mathrm{C}=\mathrm{q} / \mathrm{V}
$$
Where,
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It works on principles of Coulomb’s law in which like charges repel each other while unlike charges attracts each other thus unlike charge polarity gets induced on the inside surfaces of charged plates of a capacitor. The conductor thus holds equal and opposite charges inside the surface and thus the formation of electric field inside the non-conducting region.
Here's a simple breakdown of how Capacitor works:
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Capacitors not only store charge but store energy as well thus when it is connected to the circuit there is decrease in charge as well as transfer of charge takes place.
Consider a parallel plate capacitor. Let C be the capacitance at time t. Charge on the capacitor and the potential difference between capacitor plates is zero. Potential difference increases when the charge is given. Then potential difference between its plate is V=q/C where q is the charge at any instant.
Work done in providing more charge is given by:
$$
\begin{aligned}
& \mathrm{dW}=\mathrm{Vdq} \\
& \mathrm{dW}=\mathrm{dq} \mathrm{q} / \mathrm{C}
\end{aligned}
$$
On integrating the equation from 0 to 1 , $\mathrm{E}=\mathrm{Q}^2 / 2 \mathrm{C} \quad$ (Work Done)
We know that charge, $\mathrm{Q}=\mathrm{CV}$
Total energy stored in the capacitor,
$$
\begin{aligned}
& \mathrm{E}=\mathrm{Q}^2 / 2 \mathrm{C} \\
& \mathrm{E}=\mathrm{CV}^2 / 2 \\
& \mathrm{E}=\mathrm{QV} / 2
\end{aligned}
$$
Work done by the capacitor to accumulate charge inside it is equal to Energy stored inside the capacitor.
Or it is given by
$\mathrm{W}=\mathrm{Cv}^2 / 2$
We can determine Power of the capacitor by multiplying the voltage (V) across terminals and current(I), or
$\mathrm{P}=\mathrm{VI}$
As we know, E=Q2/2c, thus E∝Q2
Hence on doubling the value of Q, Energy stored
becomes 4 times. Hence the correct option is (3).
1 Both Electric potential energy and potential difference decreases.
2 Electric potential remains the same and potential difference increases.
3 Electric Potential remains the same and the potential difference decreases.
4 Both Electric Potential energy and Potential difference increases.
We know, when the capacitor is disconnected then there is flow of charge and Electric Potential Energy remains the same , as charge keeps on decreasing so does potential difference. Hence the correct answer is option(3).
Capacitor energy is the work by which the capacitor gets charged.
Electrical potential energy
Capacitance, C=q/V
Where, q= charge develop on each plate
V= voltage between them
Coulomb
Mica capacitor, film capacitor, ceramic capacitor, paper capacitor etc.
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