Transcript Physics_A2_28_PointCharges

Book Reference : Pages 86-88
1.
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....
Point charge Q & test
charge q
1.
2.
The separation of the objects is
much greater than the size of
the object
If its charge does not affect the
electric field it is in
+Q
r
+q
This is comparable to assumptions
made about the separation and
diameter of planets during gravitation
Coulomb’s law gives us the force :
F = 1 Qq
40 r2
By definition the electric field strength (E=
F/q) making the electric field strength at a
distance r from Q
E= 1 Q
40 r2
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 10-12 F/m
e = 1.6 x 10-19 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 :
F1
-Q1 point charge
F2
Test charge +q
+Q2 point charge
Our test charge experiences two forces F1 = qE1
& F2 = qE2 The resultant F is simply F = F1 + F2
The resultant field strength E = F/q = (qE1 + qE2) /q
E = E1 + E2
2. Forces in the opposite direction :
F1
+Q1 point charge
F2
Test charge +q
+Q2 point charge
Our test charge experiences two forces F1 = qE1
& F2 = qE2 The resultant F is simply F = F1 - F2
The resultant field strength E = F/q = (qE1 - qE2) /q
E = E 1 - E2
3. Forces at right angles:
Test charge +q
F1
F2
+Q1 point charge
-Q2 point charge
Standard resolving techniques... From
Pythagoras F2 = F12 + F22
Electric Field Strength E2 = E12 + E22
Trigonometry can be used to find the resultant direction
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
+Q
Y
X
+600 V
+1000 V
+400V
Ep = QV
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
at 1000V Ep = 2x10-6 x 1000 = 2x10-3J
at 400V Ep = 2x10-6 x 400 = 8x10-4J
The change in potential energy is 1.2x10-3J
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
-ve plate
charged parallel plates)
then the equipotentials are 0
equally spaced and parallel
to the plates
Equipotential Lines
+ve plate
+V
Potential V
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)
Distance d