Section v 17 Electric Fields

7/28/2019 Section v 17 Electric Fields

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Electric Fields

17. Electric Fields

Content17.1 Concept of an electric field

17.2 Uniform electric fields

17.3 Force between point charges

17.4 Electric field of a point charge

17.5 Electric potential

Learning Outcomes

(a) show an understanding of the concept of an electric field as an example of afield of force and define electric field strength as force per unit positive charge.

(b) represent an electric field by means of field lines.

(c) recall and use E = V/d to calculate the field strength of the uniform fieldbetween charged parallel plates in terms of potential difference and separation.

(d) calculate the forces on charges in uniform electric fields.

(e) describe the effect of a uniform electric field on the motion of chargedparticles.


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* (f) recall and use Coulomb's law in the form F = Q1Q2/4or2 for

the force between two point charges in free space or air. * (g) recall and use E = Q/4or

2 for the field strength of a pointcharge in free space or air.

(h) define potential at a point in terms of the work done in bringingunit positive charge from infinity to the point.

(i) state that the field strength of the field at a point is numericallyequal to the potential gradient at that point.

* (j) use the equation V = Q/4or for the potential in the field of apoint charge.

(k) recognise the analogy between certain qualitative andquantitative aspects of electric field and gravitational fields.


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Forces between charges,

Coulombs Law French scientist Charles Coulomb on investigating the force

between charges discovered that :

The force is proportional to the product of the chargesand inversely proportional to the square of the distancebetween them i.e F = kQ1Q2/r

2 where k is a constant of

proportionality, the value of which depends on the medium

around the charges Since the variation of the force is proportional to 1/r2, it is often

referred to as an inverse square law

Strictly speaking this law applies to point charges but can beused for charged spheres provided their radii are small

compared to their separation

k = 0 where 0 is called the permittivity of free space (i.e

charges are in a vacuum) given by 8.85 x 10-12 C2 N-1 m-2 (or F

m-1)

k = 8.99 x 109 C2 N-1 m-2 (or F m-1)

Permittivity in air is 1.00050 i.e. close to a vacuum


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Example

A charge of 2.0 x 10-8 C is at a distance of 5 cm from

another charge of - 5.0 x 10-8 C. Calculate the magnitude

of the force between the charges and state its nature.

(o = 8.85 x 10-12 Fm-1)

Solution

F = kQ1Q2/r2= - 3.6 x 10-3 N (attractive since minus sign)


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Concept of Electric Fields

Electric charges exert forces on each other when they are adistance apart

The idea of an electric field, which is a region in space where a

stationary charge experiences a force, is used to explain this

action at a distance

Each charge somehow modifies the properties of the spacearound itself, creating an electric field

The direction of an electric field is defined as the directionin which a positive charge would move if it were free to doso


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Electric fields

An electric field can be caused either by a point charged particle orthrough the maintenance of a potential difference between two

parallel plates.

The former is a radial field while the latter a uniform field.

For any electric field the lines of force start on a positive charge and

end on a negative charge

The lines of force are smooth curves which never touch or cross

The strength of the field is indicated by the closeness of the lines:

the closer the lines, the stronger the field

For point charges the field lines appear to come from or go into the

centre For a radial field the lines of force radiate outwardly from a positive

point source but inwardly towards a negative source.

In a uniform field the direction of the lines of force is from the

positive plate towards the negative plate.


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Electric field strength, E

Theelectric field strength E, is defined as the forceper unit charge acting on a small positive chargeplaced at that point,

i.e. E = F/Q N C-1

The field lines of such fields are parallel to each otherand are spaced at even distances from each other.

Uniformity of the field is only at the centre between theplates and not at the edges.


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Potential difference, V and uniform

electric field

The potential difference VABbetween two points A and B in anelectric field is the work done W, in moving a unit positivecharge from B of lower potential to A of higher potential against

the direction of the line of action of the force on the unit charge.

That is VAB = W/Q i.e. W = VQ SI unit of V is Volts

But also, work done (which is energy) W = Fd, therefore Fd = VQwith units ofV C

Rearranging, F/Q = V/d,

but F/Q is the force per unit charge which is the definition of

electric field strength E. Hence for a uniform field, the field strength E = V/d V m-1

E has 2 equivalent SI units, V m-1 = N C-1


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Example

Two metal plates 5.0 cm apart have a potential difference of 1000 V

between them. Calculate:

(a)The strength of the electric field between the plates

(b)The force on a charge of 5.0 nC between the plates

Solution

(a)E = V/d = 2.0 X 104 V m-1

(b)F = EQ = 1.0 x 10-4 N


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Field strength due to an isolated point

charge

The field pattern due to an isolated point charge is radial and, if it is

not isolated, if any other object charged or otherwise is near it, the

field will be distorted

From Coulombs Law, the force on a test charge q, a distance r fromthe isolated point charge Q is given by, F = Qq/4

0r2

The electric field E at the location of the test charge q, is given by

E = F/q

Therefore the electric field due to the isolated point charge is

E = Q/40r2 i.e. inverse square law because the force

varies in proportion to 1/r2

or E = kQ/r2 where k = 1/40 = 8.99 x 109 C2 N-1 m-2 (or F m-1)


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Example

In a simplified model of the hydrogen atom, the electron is at a distance of

5.3 x 10-11 m from the proton. The proton charge is 1.6 x 10-19 C. Calculate

the electric field strength of the proton at this distance.

Solution

Assuming that the field is radial,E = kQ/r2= 8.99 x 109 x 1.6 x 10-19/(5.3 x 10-11)2

= 5.1 x 1011 N C-1


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Electric potential energy and electric potential

Energy is a conserved quantity and is thus a fundamental aspect of nature

We can use the idea of energy in electricity also and define electricpotential energy in the same way as other types of potential energies

The change in electric potential energy when a charge Q is moved between2 points A and B in an electric field is the work done by the electric force

in moving the charge from B back to A Electric field has been defined as the force per unit charge

Electric potential V, at a point in an electric field is defined as the workdone or potential energy PE, in bringing unit positive charge frominfinity to the point i.e V = PE/Q

In dealing with gravitational potential energy we take the floor or theearths surface as zero in computing mgh. In electrical problems the earths

potential is taken to be zero although the official definition of the potentialis the potential of a point an infinite distance away


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Relationship between electric field at a point and

electric potential at that point

The field strength is equal to the negative of the potential gradient at thatpoint

V = PE/Q = (F x r)/q = (Qq/(40r2) x r)/q = Q/40r

= kQ/r

where k = 1/40r


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Example

How much work must be done by an external force in moving a charge qof+2.0 C from infinity to a point A, 0.40 m from a charge Q of +30 C?

Solution

The work is simply the change in electric potential energy. The potential

energy at infinity is zero, soW = qV = qkQ/r = 1.4 J

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Section v 17 Electric Fields