Book Reference : Pages To understand what we mean by “point charge” 2.To consider field strength as a vector 3.To apply our knowledge of equipotentials to electric fields 4. To understand what is meant by potential gradients

We can consider a charge to be a “point charge” if The separation of the objects is much greater than the size of the object 2.If its charge does not affect the electric field it is in This is comparable to assumptions made about the separation and diameter of planets during gravitation +Q +q r Point charge Q & test charge q

Coulomb’s law gives us the force : F = 1Qq 4  0 r 2 By definition the electric field strength (E= F/q) making the electric field strength at a distance r from Q E = 1Q 4  0 r 2 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 away from a nucleus with a charge of +82e  0 = 8.85 x F/m e = 1.6 x C

If our test charge is in an electric field due to multiple charges each exerts a force. The resultant force per unit charge (F/q) gives the resultant field strength at the particular position of our test charge We can consider 3 scenarios:

1. Forces in the same direction : Our test charge experiences two forces F 1 = qE 1 & F 2 = qE 2 The resultant F is simply F = F 1 + F 2 The resultant field strength E = F/q = (qE 1 + qE 2 ) /q E = E 1 + E 2 Test charge +q+Q 2 point charge-Q 1 point charge F1F1 F2F2

2. Forces in the opposite direction : Our test charge experiences two forces F 1 = qE 1 & F 2 = qE 2 The resultant F is simply F = F 1 - F 2 The resultant field strength E = F/q = (qE 1 - qE 2 ) /q E = E 1 - E 2 Test charge +q+Q 2 point charge+Q 1 point charge F1F1 F2F2

3. Forces at right angles: Standard resolving techniques... From Pythagoras F 2 = F F 2 2  Electric Field Strength E 2 = E E 2 2 Trigonometry can be used to find the resultant direction Test charge +q -Q 2 point charge+Q 1 point charge F1F1 F2F2

Equipotentials are lines of constant potential & can be compared to contour lines on a map. (and are the same as we have encountered for gravitation) +Q A test charge moving along an equipotential has constant potential energy & so no work is done by the electric field Equipotential lines and field lines always meet at right angles

Note the lines of equal potential (measured in V) are shown by the equipotential lines Consider the change in potential energy if a test charge of 2  C is moved from X to Y +Q V +400V +600 V X Y E p = QVat 1000V E p = 2x10 -6 x 1000 = 2x10 -3 J at 400V E p = 2x10 -6 x 400 = 8x10 -4 J  The change in potential energy is 1.2x10 -3 J

Definition : The potential gradient is the change in potential per unit change in distance in a given direction Two scenarios: Non uniform field : The potential gradient varies according to position & direction. The closer the equipotentials the greater the potential gradient

Uniform field : When the field is uniform, (e.g. Between oppositely charged parallel plates) then the equipotentials are equally spaced and parallel to the plates Graph shows that potential relative to the –ve plate is proportional to distance (pg is constant & is V/d) (Potential increases opposite direction to field) Equipotential Lines -ve plate+ve plate Potential V Distance d 0+V