Electric Field SS1 Physics Lesson Note

An electric field is a region of space which surrounds a system of electric charges. Electrical forces will act on any electric charge which is placed within the region.  The electric field is a vector quantity.  The direction of the field can be determined using a test charge (a small positive charge)

Fundamental Law of Electrostatics

The fundamental law of electrostatic states that: “Like charge repels, unlike charges attract.

COULOMB’S LAW

Coulomb’s law states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.

Mathematically,

F α q1q2

        r2

F =k q1q2

        r2

Where k = 1 = 8.99×109

                4πɛ◦

Thus,

F = q1q2

      4πɛ◦r2

Electric Field Intensity Or Strength (E) 

The electric field intensity, E, at any point in an electric field, is the force experienced by a unit positive test charge at that point.  It is a vector quantity whose S. I unit is (N/C), mathematically.

E = F ÷ Q

E= Electric field intensity (NC-1); F = Force, q = charge.

Also, F between Q and q is given as 

F = Qq

      4πɛ◦r2

But E = F  =    Qq x     1

            Q     4πɛ◦r2 q

  :. E = Q

       4πɛ◦r2

Electric Potential 

The electric potential (V) at a point is the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field.  It is measured in volts.  It is a scalar quantity.

Mathematically, v = w ÷ q

where V= electric potential (volts); W= work done in joules; q = charge in coulombs

The electric potential at a point due to a charge Q at a distance r from the charge Q at a distance r from the charge is given as:

V = Q

     4πɛ◦r

If the work done is against the field, the potential is positive.  If the work done is by the field, the potential is negative.  The potential of an infinity is zero.  Also, the potential of the earth is zero.  The earth is used to test the potential of the body.  This is done by connecting a wire from the body to the earth (the body is said to be earthed).  If electrons flow from the body to the earth, the body is at a negative potential.  If an electron flows from the earth to the body, the body is at positive potential.  Positive points are at higher potential while negative points are at lower potential.

Potential Difference 

The potential difference between any two points in an electric field is the work done is taking a unit positive charge from one point to another in the field.

If a charge Q is moved from a point at a potential V1 to another at a potential V2, the potential difference (V1 – V2) is the work done by the field.

Work done on the charge, 

W = Q (V1 – V2)

If Q moves from A to B, then the work done,

W = Force x distance

W = F.X

But E = F

Q

EQ = F

: .W = EQ.X

But W = Q(Va – Vb)

        Q (Va – Vb) = EQ. X

Va – Vb = E X

E =  Va – Vb

X

:.  E =p.d.

       distance

i.e, E = V

            X

V = E X

: .  V =Q

         4πɛ◦

Electron Volt (eV)

The electron volt is the quantity of energy gained by an electron in accelerating through a potential difference of one volt.

Electronic charge = 1.6 x 10-19C

I e V = 1.6 x 10-19 x 1 = 1.6 x 10-19J

The energy acquired by a charged particle accelerated by an electronic field in a vacuum depends only on its charge and the p.d. through which it falls.  When the electron is in motion, its kinetic energy will be ½ mv2.  If the electron moves in a circle of radius r, the force towards the centre in mv2 (centripetal force), and is provided by the electrical force of attraction.

Force of attraction   =     e2

                                     4πɛ◦r2

: . ½ mv2 =     e2

                    4πɛ◦r2

= 1/2{  e2 ÷ 4πɛ◦r2}

EXAMPLE

1. Calculate the energy in eV and Joule of an α particle (helium nucleus) accelerated through a p.d. of 4 x 106V.

Solution 

The charge on an α particle is 2e.

KE = work done

KE = charge x p.d. = q x v

= 2 x 4 x10^6

= 8 x 10^6 eV = 8 MeV

1eV = 1.6 x 10^-19J.

KE gained = 8 x 10^6 x 1.6 x 10^-19

= 1.48 x 10^-12J

CAPACITORS AND CAPACITANCE

CAPACITORS

A Capacitor is a device for storing electrical energy or charges. In general, capacitors can be in the form of two conductors which are insulated electrically from the surroundings. 

However, the most common types of capacitors are in the form of two parallel plate conductors which are separated by a very small distance, d. The two plates of the capacitor can be made to carry equal and opposite charges by connecting the capacitor across the terminals of a battery such that the potential difference across the plate is V.

The capacitor is represented as —–| |—-

CAPACITANCE

The capacitance of a capacitor is defined as the ratio of the charge Q on either conductor to the potential difference V between the two conductors

C = Q/V

Q = CV

The SI unit of capacitance is the farad (F) which is equivalent to coulomb per volts (CV-1)

Factors that affect the capacitance of a capacitor are:

1. The area of the plates
2. The separation between the plates
3. The di-electric substance between the plates

For a parallel plate capacitor, the capacitance C is given by

C =ɛA

      d

Where:

An area of the plates

d= their separation

ɛ= permittivity of the dielectric medium (Fm-1) 

CAPACITOR IN SERIES AND PARALLEL

If two or more capacitors are c1, c2 … are connected in series, it can be shown that the equivalent of net capacitance, c of the combination is given by:

1/c = 1/c1 + 1/c2 +…

If they are connected in parallel the net capacitance C in this is given by:

C = c1 + c2 + …

Note: that the opposite is the case if these were resistance.

Examples

1. A capacitor contains a charge of 4 .0 x 10– 4 coulomb when a potential difference of 400 v is applied across it. Calculate the capacitance of the capacitor.

Solution:

The capacitance C = q/v

= 4.0 x 10^-4

      400

= 10^ – 6F

= 1.0f

Energy Stored In Capacitor 

A charge is a store of electrical energy. When a charge, q, is moved through a p.d, the work done is given by 

       W = average p.d x charge

          = ½ qv = ½ QV

      But v = q /c; V = Q

                                   C

W= ½ q/c x q = ½ q2/c

W = ½ Q/C

Using Q=CV

W =1/2CV2

Therefore the work done is either 

¼ q2/c or 1/2cv2  = W = ½ CV2

This work is stored in the capacitor as electrical potential energy