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# Capacitor And Capacitance Assignment Help

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Electrostatics - Capacitor And Capacitance

**Physics Assignment Help >> Electrostatics >> Capacitor And Capacitance**

**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 **Q**_{1} = + Q be charge on one conductor,

Whose potential is **V**_{1} and Q_{2} = - Q be the charge on second conductor whose potential is **V**_{2} net charge on the capacitor = **Q**_{1} + Q_{2} = + Q – Q = 0

For calculation charge on the capacitor is taken as **Q** and potential difference between the two conductors is **V = V**_{1} – V_{2}

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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**Physics Assignment Help >> Electrostatics >> Capacitor And Capacitance**

**Capacitor And Capacitance**

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

**Q**be charge on one conductor,

_{1}= + QWhose potential is

**V**be the charge on second conductor whose potential is

_{1}and Q_{2}= - Q**V**

_{2}net charge on the capacitor =

**Q**

_{1}+ Q_{2}= + Q – Q = 0For calculation charge on the capacitor is taken as

**Q**and potential difference between the two conductors is

**V = V**

_{1}– V_{2}

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

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

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}farad1 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**

Q = CV = 10

^{ -12}F = 10^{-11}FQ = CV = 10

^{– 11}x 250 = 2.5 x 10^{–9}C

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