+1-415-670-9189

info@expertsmind.com

# Capacitor And Capacitance Assignment Help

###
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

**Electrostatics Assignment Help - Live Physics Tutors 24x7 Hrs**

Are you struggling with electrostatics physics problems? Electrostatics questions are giving you trouble? Do not need to worry about your subject; ExpertsMind.com makes easy electrostatics theory and problems by giving you conceptual and tricky approach for solving your complex subject problems. ExpertsMind.com offers Electrostatics Assignment Help, Electrostatics Homework Help and Physics Question's Answers with best approach for solving particular problem which makes easy to solve same kind of questions in future.

**Popular Tags: ****Capacitor And Capacitance Assignment Help, Capacitor And Capacitance Homework Help, Capacitor And Capacitance Projects Help, Online Tutoring, Capacitor And Capacitance Based Assignments Help, Capacitor And Capacitance tutors Online**

**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

**Electrostatics Assignment Help - Live Physics Tutors 24x7 Hrs**

**Popular Tags:****Capacitor And Capacitance Assignment Help, Capacitor And Capacitance Homework Help, Capacitor And Capacitance Projects Help, Online Tutoring, Capacitor And Capacitance Based Assignments Help, Capacitor And Capacitance tutors Online**