Capacitor And Capacitance Assignment Help

Electrostatics - Capacitor And Capacitance

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Capacitor And Capacitance

A capacitor is an arrangement for storing large amounts of electric charge and hence electrical energy in a small space. 

Usually a capacitor consists of a system of two conductors separated by an insulating medium often, the two conductors are charged by connecting them to the terminals of a battery.

Let Q1 = + Q be charge on one conductor, 

Whose potential is V1 and Q2 = - Q be the charge on second conductor whose potential is V2 net charge on the capacitor = Q1 + Q2 = + Q – Q = 0

For calculation charge on the capacitor is taken as Q and potential difference between the two conductors is V = V1 – V2

We know that electric field in the region between the conductors is directly proportional to charge Q. the potential difference V = work done per unit positive charge in taking small test charge form conductor 2 to conductor 1 against the electric hence the ratio Q/V must be constant.

Q/V = constant = C

The constant C is called capacitance of the capacitor. 

We may define capacitance of a capacitor as the ratio of charge on the capacitor to the potential of the capacitor. 

Capacity of a capacitor depends neither on Q nor on V. It depends on geometrical configuration (shape size separation) of the system of two conductors. C also depends on nature of insulating medium separating the two conductors. However C does not depend upon nature of material of the conductors. 

If V = 1 then from eq. (1)

Q = C

C = Q

Hence capacitance of a capacitor is the amount of charge required to raise its potential by unity. 

The SI unit of C is farad. If Q = I coulomb and V = 1 volt, then from (1) C = 1 farad.

Capacitance of a conductor is said to be one farad when a charge of one coulomb raises its potential through one volt. 

Smaller units of capacitance are

1 micro farad 1μ F = 10-6 farad,

1 nanofarad 1nF = 10-9 farad.

1 micro microfarad 1 μ μF = 10–12 farad.

μμ F is also called picofarad (pF)

A potential difference of 250 volt is appalled across the plates of a capacitor of 10 pF. Calculate the charge on the plated of the capacitor.

Sol here, V = 250 V,

C = 10p F = 10 x 10 -12 F = 10-11 F

Q = CV = 10 – 11 x 250 = 2.5 x 10–9 C


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