Transcript Work done by electric force (source: fixed charges) on a test charge

Work done by electric force (source: fixed charges)
on a test charge
depends only on endpoints, not on path.
(You can see this easily for a single fixed charge… it holds in general
because of superposition.)
Electric forces are “conservative” - We can define a potential energy.
When a + charge moves “down the field”, the electric force does work
on it, increasing its kinetic energy (or putting energy elsewhere).
When a + charge moves “up the field”, it either loses kinetic energy, or
some other force must push it up.
Example : a charge compressing a spring.
The electrical potential energy of a system of charges is the work
necessary to assemble the charges from “infinity”.
(For point charges, we take U=0 at infinity.)
This will include all pairs of interactions.
Two equal + charges are initially stationary and separated by r0.
If they are allowed to fly apart (to infinity), what will be the kinetic energy
of each?
q
A] 40 r0
q2
B] 40 r0 2

1 q2
C]
4 0 
r0
1 q2
D]
8 0 r0


1
1
Three equal + charges are initially stationary and at the vertices of an
equilateral triangle with side r0.
If they are allowed to fly apart (to infinity), what will be the kinetic energy
of each?
q
A] 40 r0
q2
B] 40 r0 2
1 q2
C]
4 0 
r0
1 q2
D]
8 0 r0
1

1

Just as we can define electric field as the force felt by a test charge
We define “potential” as potential energy of a test charge.
1 q pc
Vpc 
40 r
dV 
1 dq
40 r
Just as a conservative force is:
(minus) the derivative of the potential energy

 W  U   dU 
The electric field is
(minus) the derivative of the potential.

 Fdx
Note: just because V=0 doesn’t mean E=0!
A function can be zero but have a non-zero derivative.
Also: it’s time to think in 3D. The derivative can be taken w.r.t
x, y, or z.
dV  E  dl
dV  E x dx  E y dy  E z dz
V
Ex  
x
E  V
This means: hold y, z constant, so dy=dz=0
Note that E is a vector, but V is a scalar.
Rank the magnitudes (smallest to
largest) of the electric field at
point P in the three arrangements
shown. ANSWER=D
A] all are the same
B] I, II, III
C] III, II, I
D] II, I, III
Rank the electric potentials at
point P (smallest to largest).
ANSWER=A
If V = -4x2 + 4y (x,y in meters, V in volts) , what is the E field at the
origin?
ANSWER=C
A] 0
B] 4 V/m in the +y direction
C] 4 V/m in the -y direction
D] 4 V/m directed at an angle between +x and -y axes.
E] cannot determine
If V = -4x2 + 4y (x,y in meters, V in volts) , where is the E field = 0?
ANSWER=E
A] at the origin
B] at x=1, y=1
C] at x=-1, y=1
D] at both B and C
E] nowhere.
Equipotential Lines = Contours of constant V
E field
points downhill
Downhill is
always
perpendicular to
level
Conductors at
rest are
equipotential