Download Ch19: Electric Potential Work done transfers energy Let`s work

Think, Answer; Talk to your neighbors, Answer
Ch19: Electric Potential
A positively charged particle pushed up at constant
speed by thin wire a distance d as shown in the
animation..
While moving up,
the sign of the net work done is ?
Or is the work done zero?
A: +
B: C: zero
D: Not enough info
d
Work done transfers energy
Fyou
x
x
Fgravity
You do the work against
the gravitational field. The
work gets transferred into
change in Gravitational
potential energy.
++++++++++++++++++++ Q
E
q
-----------------
The wire does work
against the Electric field.
The work gets transferred
into change in Electric
potential energy. Work by E:
WE  FE x cos 
U  WExtF  WFieldF
 qE x cos180o   qE x
U E  ?
Think, Answer; Talk to your neighbors, Answer
An external force that is not shown is making the
charge move with constant velocity. The work done
by the agent, done by the electric field, and done by
the net force on the charge +q are:
Agent
E-Field
Net
Force
A
+
-
+
C
-
+
0
B
D
-
+
+
-
-
0
E: None of these/I wish I knew
E
+q
Recall, work
=
Force x displacement
x cos(angle between them)
++++++++++++++++++++

-----------------
Let’s work together
Q
1. Draw the direction of electric field
+++++++++++++
between two parallel charged plate.
x
2. What is the direction of force on the E
e
negative charge? Draw it.
-----------
3. What is the work done by electric force on the charge?
4. What is the change in electric potential energy?
From last chapter, E due to charge on parallel plate: E 
Q
o A
5. What is the change in the PE of a proton when it moves
2.0 cm? Assume that each square shaped plate has side
length 1.0 cm and charge on the plate is 8.85 µC.
PE of point charges
Any two charges separated by r.
Qq
FE  k 2
r
.
Q
r
.
q
If r is very-very large (), FE = 0. No work is required to
move the charge around over there means UE (at )= 0
When r is finite FE ≠ 0. Work done to bring there turns out
to be
Qq
Qq
r k
r
r2
Since, U  WE
it becomes:
Qq what happened to the sign?
U f  Ui  k
r
WE  FE r  k
It turns out, using the sign of charges takes care of the
sign of the PE.
Qq
Since Ui = 0, U E (orU f )  k
r
1
.
Problem: What is the potential energy if Q = 2C , q
= 1 C and r = 9 cm?
r
Q
.
Electric Potential (or voltage)
Electric Potential energy per unit charge is voltage: V 
q
Problem: How much work does it take for a external force to
set up the arrangement of charged objects in the diagram on
the corners of a right triangle when the three objects are
+
q1  5.5  C
initially very far away from each other.
The work done on the charges is
equal to their potential energy.
qq q q qq 
UE  k  1 2  2 3  1 3 
r23
r13 
 r12
r12 = 12 cm
-
r23 = 16 cm
+
q2  6.5  C q3  2.5  C
ans : 3.0 J
UE
q
Just like Field is force per unit charge.
Unit of Potential: J/C called Volts and written as V.
Notice both the potential and its unit are denoted by V. 
Just like a field tells us about the force on a test charge
around a source charge, the potential tells us about the
potential energy of the test charge.
U E  qV
k
Qq
 qV
r
V 
kQ
r
Change in potential between two
points is called Potential difference.
U E
V 
q
or, U E  qV
.
r
V
++
++
Source
charge, Q
Think, Answer; Talk to your neighbors, Answer
ECG is display of change in Potential with time
Two identical charges, +Q and +Q, are fixed in
space. What is the magnitude of the E field, and the
value of the voltage, at the midpoint between them?
(Assume the potential is zero at infinity.)
+Q
E=?
V=?
(Midpoint)
+Q
A) E=0 , V nonzero B) E nonzero , V=0
C) Both are 0
D) Both are nonzero
E) Not enough information
Making sense of a battery Label
Think, Answer; Talk to your neighbors, Answer
Problem: If the 9V battery lights 12 W bulb, how
many electrons are flowing through the bulb each
second?
Did the pot. energy (PE) increase or decrease?
Did the voltage (V) at the position of the test charge
increase or decrease?
What does it mean when it says 9V on
this battery?
A negative charge -q is released from position i to
position f between the charged plate.
A: PE , V
B: PE , V
C: PE , V
D: PE , V
E: None of these.
Problem: What is the potential difference between
the two parallel plates separated by distance d?
Assume charge and area of each plate is Q and A.
V 
U E
qE x
 Ed

q
q

Q
d
o A
+++++++++++++
.
.
-----------
+++++++++++++
E
f
i
q
d
-----------
2
Where is higher PE? Potential
To decide higher PE, let +q move.
+++++++++++++
And think about how rocks roll
down hill. High hill, high gPE.
E
q
-----------
Similarly, +q moves from higher
PE to lower PE. How about -q ?
Now how to find where is the Potential high or low?
E points from high V to low V (always!).
Is ∆UE of the particle positive, negative, or zero as the
particle moves from i to f.
We know: U  U  U
E
f
i
e
e
+
i
f
Just like one might choose sea level, or the tabletop, or
the ground as zero gPE, you can pick any spot you want
and call the electrical potential energy 0 there. Same is
true for potential.
Potential and field due to
a spherical conductor
E
E0
inside
1
r
2
outside
This is field due to sphere.
V
constant
Potential due to the sphere.
Notice the potential difference
between two consecutive
equipotentials is same.
V 
or, E 
V
d
q

q
e
b
c
Q
Q
r
a
g
At which labeled point is voltage highest?
B
A
D
E
 Ed
means, E and d are inversely proportional.
Also remember closer to the
charge, field is stronger.
V k
Think, Answer; Talk to your neighbors, Answer
Equipotential and Field
But the equipotentials are
not uniformly distributed.
Recall:
U E
qE x
Are points a, b, c, d at same
distance from the charge Q?
Is potential, V, same at
these points?
It is equipotential circle
for 2D and surface for
f
3D (sphere).
d
Which circle has larger
potential?
Do we know how to draw E
filed lines due to Q?
Geometrically, how are they related? draw E field
lines before you answer.
1
r
outside
inside
Equipotential Surfaces represents V field
Ek
Q
r2
So E is stronger in the region where equpotentials are
crowded.
Equipotential
Field lines
3
Electric Potential and Conservation of energy
Ei  E f
or, K i  U i  K f  U f
Problem: A proton is
released from rest at
point a. It then travels
past b. What is its speed
at b?
KE is our good old
PE is qV
1 2
mv
2
Problem: A proton is fired toward a small hole in the
negative plate with the speed shown. What is its kinetic
energy when it emerges through the hole in the positive
plate? (Hint: The electric field outside of a parallel-plate
capacitor is zero).
Note: Small energy is expressed as electron volt, eV.
1eV  1.6  1019 J Does this number look familiar?
Think, Answer; Talk to your neighbors, Answer
Capacitors
Q  C V
A
where,C  o
d
If we double L (both of them!) and halve d, by
what factor have we changed capacitance?
When two metal plates
separated by air or insulator
makes a capacitor.
Simplified sketch:
L
Q
U E qE x
Q
+++++++++++++
V 
d

 Ed 
d
E
q
o A
q
o A
----------or, Q 
 V  C V
 A
d
where,C  o
It is called capacitance.
d
Charging a Capacitor
When you throw a ball
higher, it gains gPE.
Ball at height has energy.
---+++++
Battery separates the
charges on the plates.
+++++++++++++
L
d
A: no change
D: up by 8
B: up by 2.
C: up by 4.
E: none of these
Work, Answer; Talk to your neighbors, Answer
A capacitor charged to 1.5 V stores 2.0 mJ of energy. If the
capacitor is charged to 3.0 V, it will store
A.
B.
C.
D.
E.
1.0 mJ
2.0 mJ
4.0 mJ
6.0 mJ
8.0 mJ
1
U  Q V
2

Q2
2C
1
 C V 2
2
----------2
1
Energy supplied by the battery is stored
U  Q V  Q
as electric energy in the capacitor.
2C
2
4
Problem: A 12 V car battery is hooked up to a capacitor
with plates of area 0.1 m2, a distance 8.85 mm apart.
(a) How much charge builds up on a plate?
Dielectrics in a Capacitor
Vacuum or air
(b)How much energy is stored in the capacitor now?
The electric field due to polarized dielectric reduces net
electric field and hence lowers the V . Recall V =Ed.
Since, C 
Q
, small V means larger C.
V
So capacitance increase by a factor called dielectric
 A
constant, .
Cdiel   o   C
d
Problem: A capacitor can be made from two sheets of
aluminum foil separated by a sheet of waxed paper. If
the sheets of aluminum are 0.3 m by 0.4 m and the
waxed paper, of slightly larger dimensions, is of thickness
0.030 mm and has  = 2.5, what is the capacitance of this
capacitor?
ans 8.85*10^8 F
Problem: A parallel-plate capacitor is charged using a
100 V battery; then the battery is removed. If a
dielectric slab is slid between the plates, filling the
space inside, the capacitor voltage drops to 30 V.
What is the dielectric constant of the dielectric?
ans 3.3
5