ELECTROSTATICS

An object becomes electronically charged when the number of protons and electrons
do not balance.

Static electricity results when this charge is isolated by insulating material.

Excess electrons – object will be charged negatively.

Deficit of electrons – objects will be positively charged.

LIKE CHARGES REPELUNLIKE CHARGES ATTRACT

When one object is rubbed by another electrons are often transferred. If a one or both objects are insulators or insulated then static electricity results.

e.g polythene rubbed with flannel becomes negative.

e.g. glass rubbed with silk becomes positive.

ELECTRIC FIELDS

An electrically charged object always generates an electric field around it.

Definition: An electric field is a region of space in which an electric charge experiences a force.

The electric field is continuous around an object but we use electric field lines to represent the field. These lines can be shown by placing seeds or rough flour particles in the field.

Definition: An electric field line is a line drawn in such a way that a small positively charged object placed on the line will experience a force at a tangent to the line at that point.

Properties of electric field lines:

  1. They start on positive charges and end on negative charges.
  2. They never cross.
  3. They are 3-dimensional.
  4. The denser the field lines the stronger the field.
  5. They enter and leave charges at right angles.
  6. They are continuous (although drawn as separate lines).
  1. Charged Objects:

Negatively charged object
/ Two equally positively charged objects


Hollowmetal ring
/ Non-spherical object

  1. Parallel Plates: There is a constant field between oppositely charged plates.

LAW OF CONSERVATION OF CHARGE

The algebraic sum of the electric charges in a closed system remains constant.

e.g.Two identical metal spheres, one with a charge of -2 nC and the other with a charge of +7 nC, touch one another and are then separated. What will be the charge on each sphere?

COULOMBS LAW

The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between their centers.

F α Q1.Q2F α 1/r2

k = 9 x 109 N.m2.C-2

1 nC = 1 x 10-9 C1 μC = 1 x 10-6 C

Note:All values for charges must be a multiple of the magnitude of a charge of a single electron (Qe = -1,6 x 10-19 C).

e.g. How many electrons have been removed from an object which has a charge of
+4,8 nC?

Examples:

1.An object with a +1,8 nC charge is placed 5,2 cm from an object with a charge of -8,0 nC.

a)Calculate the force the +1,8 nC object exerts on the -8,0 nC object.

b)How many excess electrons on the -8,0 nC object?

2.Two identically charged objects, placed 46 mm apart, exert a force on each other of
1,2 x 10-3 N. What is the charge on each object?

3.Two identical metal spheres, one with a charge of +8 μC and the other with a charge of
+25 μC, touch one another and are then separated to a distance of 8,0cm. What force do they exert on each other.

ELECTRIC FIELD STRENGTH

Definition: The electric field strength (E) at a point in an electric field is the force experienced per unit positive charge placed at that point.

E = F/Q

Units: N.C-1 (or V.m-1 in parallel plates)

e.g. What is the force exerted on an 8,06 nC charge in an electric field with a strength of 4000 N.C-1?

Consider a small charge q C placed in the field of a large charge Q C at a distance of r m.

Force on q:F = E.q and F = (k.Q.q)/r2

Therefore E.q = (k.Q.q)/r2

(q cancels) E = kQ/r2

e.g.What is the electric field strength at a distance of 24 mm from an object with a
charge of 6,88 x 10-14 C?

Note: If a charged object (q) is placed between two other charged objects (Qa and Qb), then each object acts independently producing the field which acts upon it.

e.gCalculate the total force acting upon the charged object X in the system belowby first calculating the resulting electric field at X.

ELECTRIC POTENTIAL ENERGY

If the object with the charge +q is moved to the left its potential energy increases because they repel. If it is moved to the right its potential energy decreases. At an infinite distance its potential energy is equal to zero.

Note: if the object with the +q charge is released and is free to move then it will accelerate to the right gaining kinetic energy as it loses potential energy.

Potential energy U of a charge q in the field of another charge Q:

F = k.Q.q

r2

If one rearranges Coulombs equation to get F.r, then F.r = (k.Q.q )/r

But W = F.Δx = F.r so W = (k.Q.q )/r

As work done on moving a charge in a field is the gain in Ep (or U) then:

U = (k.Q.q)/r

e.g. Two charges q of +1,3 x 10-16C and Q of -6,3 x 10-15 C are 7,0 cm apart. Calculate the potential energy of charge q.

U = k.Q.q = 9 x 109 x 6,3 x 10-15 x 1,3 x 10-16

r 0,07

= 1,05 x 10-19 J

Potential energy of a charge q due to a collection of charges

The potential energy of a charge due to several charges is the sum of the potential energies due to the individual charges. (Attractions and repulsions need to be taken into account.)

Point charges Q1 = +12 x 10-9 C and Q2 = -12 x 10-9 C are placed 9 cm apart, as shown in the diagram below. Calculate the potential energy of a +4 x 10-9 C point charged if it is placed at point X .

POTENTIAL DIFFERENCE (V) (voltage)

The potential difference between two points in an electric field is the work done per unit positive charge moving from a point of low potential to a point of higher potential.

V = W/Q

(V – volts, W – joules, Q – coulombs)

A potential of 1 V exists between two points if 1 J of work is required to move 1 C of charge from the point of low potential to the point of high potential.

PARALLEL PLATES

NBThe electric field between parallel plates is constant. This means a charged object experiences a constant force at any position in the field.

Consider an object with a charge of +Q between two plates a distance d apart across which a potential difference of V.

  • The object has a maximum potential energy where it is.
  • If it is free to move it will accelerate until it hits the negative plate.
  • If there is a vacuum between the plates then the kinetic energy the object gains will be equal to the potential energy it loses.
  • The work done by the field moving the object from positive plate to the negative equals W = V.Q
  • The work done by the field equals the gain in kinetic energy in a vacuum.
  • When work is done against the field a charged object gains potential energy

The electric field strength (E) between the plates can be calculated using the formula:
E = V/d
(d must be in metres.)
NB Do not mix up E with energy (Ep or Ek)! /

e.gTwo parallel plates 4,0 cm apart have a p.d. of 8000 V placed across them. A 0.5 g object with a charge of +1,2 x 10-15 C on it is held next to the positive plate. (assume no air resistance)

a)Calculate the electric field strength of the field between the plates.

The object is released.

b)Calculate the force acting on the charged object.

c)How much kinetic energy will the object have as it hits the negative plate?

d)Calculate the speed of the object as it hits the plate. (Ek = ½m.v2)

e)How much potential energy will the object have if it is at a point 1,0 cm from the positive end?