Introduction
In this chapter, we will study about Capacitor and its working. We will also study the expression for energy stored in capacitors. In the later part we will derive the equation of energy stored. In the end, we will study work done by the capacitor.
Capacitor is an electric device used to create more charge for a given voltage.
Generally combinations of capacitors are in form of:
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A capacitor is a device consisting of two equally charged conductors separated by a non-conducting region.
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 capacitor. Conductor thus holds equal and opposite charge inside the surface and thus formation of electric field inside the non-conducting region.
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SI unit of Capacitance (C) is Farad
Thus capacitance is given by the formula:
C=q/V
Where, q= charge develop on each plate
V= voltage between them
Application of Capacitor
Energy stored in a capacitor
In this section we will learn:
Note: Type of Energy stored inside the capacitor is Electric Potential energy.
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Derive the expression for energy stored in a parallel capacitor
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:
dW=Vdq=dq q/C
On integrating the equation from 0 to 1,
Work Done= Q2/2C
We know that charge, Q=CV
Total energy stored in the capacitor, U=Q2/2C =U=CV2/2=QV/2
NCERT Physics Notes:
Work done by the capacitor
Work done by the capacitor to accumulate charge inside it is equal to Energy stored inside the capacitor.
Or it is given by
W=Cv2/2
Power of the capacitor
We can determine Power of the capacitor by multiplying the voltage (V) across terminals and current(I), or
P=VI
Also check-
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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