ELECTROSTATICS AND CAPACITORS

By: Taher K D

Electrostatics• ELECTRIC FIELD: The region surrounding the charged

body in which the effect of charge is experienced is

called electrostatic field.

• ELECTRIC FLUX: The number of lines of force

emanating from a positive charge is called electric

flux. Its unit is coulomb(c).

• ELECTRIC FIELD INTENSITY: It is denoted by E. The

electric field intensity at any point in an electric field

is given by the force experienced by a positive charge

placed at that point.

• ELECTRIC FLUX DENSITY: It is denoted by symbol D and is

measured in coulomb/meter 2 .Electric flux density is

defined as the flux per unit cross sectional area emanating

normally from the surface.

• ELECTRIC POTENTIAL: It is the work done in bringing a unit

positive charge from infinity to that point against the

electric field. Its unit is volts.

• POTENTIAL DIFFERENCE: It is defined as the work done in

moving a unit positive charge within electric field from a

point of lower potential to a point of higher potential.

• POTENTIAL GRADIENT: It is defined as the rate of change of

potential or voltage with distance.

Capacitors

• Capacitor: It consists of two parallel conducting

plates separated by an insulating material called

dielectric. It is also called condenser.

• Capacitance: It is defined as the ability of a

capacitor to store electric energy in the form of

static charge. Unit-Farads.

• 1-Farad:It is defined as the capacitance which needs

a charge 1 coulomb to obtain a potential difference

of 1-volt across its plates.

Dielectric and capacitance..

•C= є0 єr A/d• Dielectric strength: It is the ability of

insulating medium to resist its breakdown when a large voltage is applied across it. Unit: V/m.

•Relative permittivity: It is defined as the ratio of electric flux density in a dielectric medium to that produced in vacuum by the same electric field strength. Unit- Farads/m

•Єr= D/D0

Storing Energy in a Capacitor

Capacitors can be used to store electrical energy. The amount of

energy stored is equal to the work done to charge it. During the

charging process, the battery does work to transfer charges from

one plate and deposit them onto the other.

Figure : Work is done by an external agent in bringing +dq from the negative plate and depositing the charge on the positive plate.

Energy storage in Capacitor

Work must be done by an external influence to "move"

charge between the conductors in a capacitor. When the

external influence is removed, the charge separation

persists in the electric field and energy is stored to be

released when the charge is allowed to return to its

equilibrium position. The work done in establishing the

electric field, and hence the amount of energy stored, is

Electric Field Energy in Capacitor

•The energy stored on a capacitor is in the form of energy density in an electric field is given by

•This can be shown to be consistent with the energy stored in a charged parallel plate capacitor

Charging a Capacitor

Consider a circuit as shown in Figure 1. Capacitor C is initially uncharged, by closing the switch S a current i is setup in the loop and the capacitor begins to charge. Applying Kirchoffs loop rule, we get

Where ε is the electromotive force (dc voltage supply), R is the resistor, Q is the charge of the capacitor and C is the capacitance. Substituting dQ/dt for the current i,Equation (1) becomes

Rearranging the terms, Equation (2) becomes

The solution of Equation (3) is given as

Cε = Q, which represents the maximum charge the capacitor can hold for a given emf. The voltage across the capacitor Vc is given as

Dividing Equation (4) by C yields

At a specific value of time t = τ = RC (called the time constant of the R-C circuit),

Therefore, by plotting Vc versus t, the time constant τ may be determined, and hence, the value of C can be calculated, provided R is known.

Equation (6) shows that the growth of the capacitors voltage is not linear, but rather grows exponentially reaching a saturation value which equals the voltage of the emfsource. The capacitor is considered to be fully charged after a period of about five time constants.The current i in the circuit at a given time t is given as

where ε/R= io represents the initial current in the circuit.Therefore, we can write

At a time t = τ , i = io e−1, or

Discharging a Capacitor

Following the same procedure as for the charging analysis, the differential equation that characterizes the discharging process is given as

The solution to Equation (13) is given as

Equation (14) determines the charge on the capacitor at as a function of time t. The voltage across the capacitor Vc :

At time t = τ = RC (the time constant), Vc becomes

The current i in the circuit is given as

where ε/R= io represents the initial current in the discharging circuit

Therefore, we can write

At time t = τ , the current i = −io e−1, or

The minus sign may be ignored and it implies that the charge is decreasing with time. Therefore by plotting i versus t, the time constant τ can be determined, and hence, the value of C can be calculated, if R is known.

The capacitor can be considered to be fully discharged, during a time lapse of five time constants.

THANK YOU