Book Reference : Pages 86-88Book Reference : Pages 86-88

1.1. To understand what we mean by “point To understand what we mean by “point charge”charge”

2.2. To consider field strength as a vectorTo consider field strength as a vector

3.3. To apply our knowledge of equipotentials to To apply our knowledge of equipotentials to electric fieldselectric fields

4.4. To understand what is meant by potential To understand what is meant by potential gradientsgradients

We can consider a charge to be a “point We can consider a charge to be a “point charge” if.... charge” if....

1.1. The separation of the objects is The separation of the objects is much greater than the size of much greater than the size of the objectthe object

2.2. If its charge does not affect the If its charge does not affect the electric field it is inelectric field it is in

This is comparable to assumptions This is comparable to assumptions made about the separation and made about the separation and diameter of planets during gravitationdiameter of planets during gravitation

+Q

+q

r

Point charge Q & test charge q

Coulomb’s law gives us the force :Coulomb’s law gives us the force :

F = F = 11 QqQq4400 rr22

By definition the electric field strength (E= By definition the electric field strength (E= F/q) making the electric field strength at a F/q) making the electric field strength at a distance r from Qdistance r from Q

E = E = 11 QQ4400 rr22

Note if Q is negative, this formula will yield negative numbers indicating that the field lines are pointing inwards

Calculate the electric field strength 0.35nm Calculate the electric field strength 0.35nm away from a nucleus with a charge of +82eaway from a nucleus with a charge of +82e

00 = 8.85 x 10 = 8.85 x 10-12-12 F/m F/me = 1.6 x 10e = 1.6 x 10-19-19 C C

If our test charge is in an electric field due If our test charge is in an electric field due to multiple charges each exerts a force.to multiple charges each exerts a force.

The resultant The resultant force per unit charge (F/q) force per unit charge (F/q) gives the resultant field strength at the gives the resultant field strength at the particular position of our test chargeparticular position of our test charge

We can consider 3 scenarios:We can consider 3 scenarios:

1. Forces in the same direction :1. Forces in the same direction :

Our test charge experiences two forces FOur test charge experiences two forces F11 = qE = qE11 & F& F22 = qE = qE2 2 The resultant F is simply F = FThe resultant F is simply F = F11 + F + F22

The resultant field strength E = F/q = (qEThe resultant field strength E = F/q = (qE11 + qE + qE22) /q) /q

E = EE = E11 + E + E22

Test charge +q +Q2 point charge-Q1 point charge

F1 F2

2. Forces in the opposite direction :2. Forces in the opposite direction :

Our test charge experiences two forces FOur test charge experiences two forces F11 = qE = qE11 & F& F22 = qE = qE2 2 The resultant F is simply F = FThe resultant F is simply F = F11 - F - F22

The resultant field strength E = F/q = (qEThe resultant field strength E = F/q = (qE11 - qE - qE22) /q) /q

E = EE = E11 - E - E22

Test charge +q +Q2 point charge+Q1 point charge

F1 F2

3. Forces at right angles:3. Forces at right angles:

Standard resolving techniques... From Standard resolving techniques... From Pythagoras FPythagoras F22 = F = F11

22 + F + F2222

Electric Field Strength EElectric Field Strength E22 = E = E1122 + E + E22

22

Trigonometry can be used to find the resultant directionTrigonometry can be used to find the resultant direction

Test charge +q

-Q2 point charge+Q1 point charge

F1

F2

Equipotentials are lines of constant Equipotentials are lines of constant potential & can be compared to contour potential & can be compared to contour lines on a map. (and are the same as we lines on a map. (and are the same as we have encountered for gravitation)have encountered for gravitation)

+Q A test charge moving along an A test charge moving along an equipotential has constant equipotential has constant potential energy & so no work is potential energy & so no work is done by the electric fielddone by the electric field

Equipotential lines and field lines Equipotential lines and field lines always meet at right anglesalways meet at right angles

Note the lines of equal Note the lines of equal potential (measured in V) potential (measured in V) are shown by the are shown by the equipotential linesequipotential lines

Consider the change in Consider the change in potential energy if a test potential energy if a test charge of 2charge of 2C is moved C is moved from X to Yfrom X to Y

+Q

+1000 V

+400V

+600 V

X

Y

EEp p = QV= QV at 1000V Eat 1000V Epp = 2x10 = 2x10-6-6 x 1000 = 2x10 x 1000 = 2x10-3-3JJat 400V Eat 400V Epp = 2x10 = 2x10-6-6 x 400 = 8x10 x 400 = 8x10-4-4J J

The change in potential energy is The change in potential energy is 1.2x101.2x10-3-3JJ

Definition : Definition : The The potential gradient potential gradient is the is the change in potential per unit change in change in potential per unit change in distance in a given direction distance in a given direction

Two scenarios: Two scenarios:

Non uniform field : Non uniform field : The potential gradient The potential gradient varies according to position & direction. The varies according to position & direction. The closer the equipotentials the greater the closer the equipotentials the greater the potential gradientpotential gradient

Uniform field : Uniform field : When the When the field is uniform, (e.g. field is uniform, (e.g. Between oppositely Between oppositely charged parallel plates) charged parallel plates) then the equipotentials are then the equipotentials are equally spaced and parallel equally spaced and parallel to the platesto the plates

Graph shows that potential Graph shows that potential relative to the –ve plate is relative to the –ve plate is proportional to distance (pg is proportional to distance (pg is constant & is V/d) (Potential constant & is V/d) (Potential increases opposite direction to increases opposite direction to field)field)

Equipotential Lines

-ve plate +ve plate

Potential V

Distance d

0 +V