Download 7TH CLASSES PHYSICS DAILY PLAN

10TH CLASSES
PHYSICS
DAILY PLAN
Ex:
SUBJECT: ELECTRIC FIELD & POTENTIAL
GOALS:
DURATION:
IN PRACTICE:
PRESENTATION:
The Electric Field : An electric field at a point in space can be
defind in terms of the negative electric force acting on a test
charge q0 placed at that point.

E
q0
q
a) Find the electrostatic force acting
on q.
370
b) If q=-1/2 C find the direction & the
magnitude of E.
q
W=12N
Electric Field Lines: A more useful way to describe an electric
field is with electric field lines, also called lines of force.
* Each line shows the path that would be followed by a free
positive test charge released in the field.
( q>>q0 )
q0

E
q0
q
Since F  k . 2
d

 F
E
q0

E
k.
E

i) E is tangent to the electric field line at each point.
ii) The field is strong where the lines are close together, it is
weaker where the lines are far apart.
q.q 0
d2 
q0
E  k.
q
d2
* The electric field is a vector quantity having the SI unit of
(N/C)
* The direction of at a point is defined to be the direction of the
electric force that would be exerted on a small positive test
charge placed at that point.
*
*
F=E.q0
+q
F
-q
F
E
Ex: q1=q
a
K
a
q3=-3q
a
q2=-3q
If the electric field produced by q1 at point K is defined
as E, then find the net electric field at K in terms of E.
* The rules for drawing electric field lines for any charge
distribution are:
i) The lines must begin on positive charges and terminate on
negative charges or at infinity in the case of an excess of charge.
ii) The number of lines drawn leaving a positive charge or
approaching a negative charge is proportional to the magnitute
of the charge.
iii) No two field lines can cross.
A
+2q
x
B
-q
Fe x  qEx
SI→ W is Joule
*
W  K  qEx  0  qEx
1
1
2
2
mVB  mV A  qEx
2
2
W E  K 
Motion of Charged Particles in a Uniform Electric Field:
E
&
(J)
(0,0)
V
V0=0
E
(x,y)
V
*
V X  V0  cons tan at
X
F=q.E=m.a

v y  at  
a
*
q.E
m
x  v0 .t

y
* If E is uniform (that is constant in magnitude &direction)
- the acceleration of the motion is constant.
- if the charge is positive the acceleration will be in the
direction of the electric field.
- if the charge is negative the acceleration will be in the
opposite of the electric field.
*
V  V0  at
V  V0  2ax
2
1 2
1 eE 2
at  
t
2
2 m
* The trajectory is parabola.
* After the electron leaves the region of uniform electric field it
continues to move in a straight line with a speed v > v0
* Notice that we have neglected the gravitational force on the
electron.
(in E=104N/C
the ratio of “elec.force/grav.force” is 10 14N
for an electron and 1011N for a proton)
1
x  x 0  V0 t  at 2
2
2
eE 2
.t
m
NOTE: The field inside a charged conductor is always zero.
Taking x0=0 & V0=0 gives;
qE 2
1
x  at 2  x 
t
2
2m
qE
t
; V  at  V 
m
(1)
(2)
 2qE 
V 2  2ax  V 2  
x
 m 
* The kinetic energy of the charge after it has moved a distance
x is;
K
1
1  2qE 
mV 2  m
 x  K  q x
2
2  m 
* Also from Work-Energy Theorem;
(2)
(4)
* The entire charge of the ball is transferred to the outer surface
of the sphere.
E
q
r
E0  k
q
r2
d
(Electric field of a charged conducting sphere as a function at
the distance from its center.)
HOMEWORK:
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DEMONSTRATION:
EXPERIMENT:
TEACHER:
DIRECTOR: