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Transcript 
5.3 Electric Potential
Work is done in
charging the
metal dome?
Why?
The van de Graaff generator
5.3 Electric Potential
Work is done in
charging the
metal dome?
Why?
‘like’ charges
are being
moved closer
to each other
The van de Graaff generator
5.3 Electric Potential
Work is done in
charging the
metal dome?
Why?
‘like’ charges
are being
moved closer
to each other
The electric potential
of the dome
is increasing
The van de Graaff generator
5.3 Electric Potential
The electric potential V at a point P in an electric field is defined as:
‘the work done per unit positive charge
on a positive small test charge when
it is moved from infinity to that point’
P
+
+
+
r
oo
5.3 Electric Potential
The electric potential V at a point P in an electric field is defined as:
‘the work done per unit positive charge
on a positive small test charge when
it is moved from infinity to that point’
Unit: V or JC-1
work done = electric potential energy gained
P
+
+
+
r
oo
Zero
Potential
energy
at oo
5.3 Electric Potential
The electric potential V at a point P in an electric field is defined as:
‘the work done per unit positive charge
on a positive small test charge when
it is moved from infinity to that point’
Unit: V or JC-1
work done = electric potential energy gained
P
+
+
+
r
oo
Zero
Potential
energy
at oo
The electric potential at point P:
Vp = Ep
Q
So electric potential energy
Ep = Vp Q
1000V
Electric
Potential
P + 1ųC
+
+
+
r
oo
Zero
Potential
energy
at oo
The electric potential at point P:
Vp = Ep
Q
So electric potential energy
Ep = Vp Q
electric potential energy
of a 1ųC test charge moving
from infinity to where the
potential is 1000V
( work done in moving
this charge from infinity
to point P)
Ep = Vp Q
Ep = 1000 x 1ųC
Ep = 1mJ
1000V
Electric
Potential
P + 1ųC
+
+
+
r
oo
Zero
Potential
energy
at oo
The electric potential at point P:
Vp = Ep
Q
So electric potential energy
Ep = Vp Q
electric potential energy
of a 1ųC test charge moving
from infinity to where the
potential is 1000V
( work done in moving
this charge from infinity
to point P)
If the test charge is moved in a field
from where the potential is V1
to a position where the potential is V2
then the work done:
ΔW = Q ( V2 -V1)
Ep = Vp Q
Ep = 1000 x 1ųC
Ep = 1mJ
1000V
Electric
Potential
P + 1ųC
+
+
+
r
oo
Zero
Potential
energy
at oo
Equipotentials: are lines joining points of constant potential.
They cross field lines at 90o.
then the work done:
ΔW = Q ( V2 -V1)
= + 2ųC (400 - 1000)
= + 2ųC x - 600
= - 1.2 x 10-3 J
Electric
potential
Energy:
Ep = Q Vp
Equipotentials: are lines joining points of constant potential.
They cross field lines at 90o.
Electric
potential
Energy:
Ep = Q Vp
Potential relative
to the negative plate
Distance X
Distance X
E = -ΔV
ΔX
Potential relative
to the negative plate
Distance X
Distance X
E = -ΔV
ΔX
Potential relative
to the negative plate
Distance X
Distance X
E = -ΔV
ΔX
Potential relative
to the negative plate
Distance X
Distance X
E = -ΔV
ΔX
Potential relative
to the negative plate
Distance X
Distance X
E = -ΔV
ΔX
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