Transcript Electrostatics Example Problems

Electrostatics Example Problems
Definition
Electric Force
Interaction
between
two
charges
Units
N
Vector / Point Charge
Scalar
Formula
vector
𝐹=𝑘
𝑞𝑠 𝑞𝑜
𝑟2
Uniform Field
Formula
𝐹 = 𝐸𝑞𝑜
Ignore signs
of qs.
Draw a
picture for
direction.
𝐹
𝐸=
𝑞𝑜
Ignore signs
of qs.
Draw a
picture for
direction.
Electric Field
𝐹
𝐸=
𝑞𝑜
Electric
Potential
Energy
Energy
that can be
converted
into kinetic
Electric
Potential
𝑉=
𝑈𝑒
𝑞𝑜
N/C
J
J/C
V
vector
scalar
scalar
𝐸=𝑘
𝑞𝑠
𝑟2
𝑈𝑒 = 𝑘
𝑉=𝑘
𝑞𝑠 𝑞𝑜
𝑟
𝑞𝑠
𝑟
Notes
Maintain
signs of qs.
∆𝑈𝑒 = −𝑞𝑜 𝐸∆𝑑
Ignore
direction.
∆𝑉 = −𝐸∆𝑑
Maintain
signs of qs.
Ignore
direction.
Electric Force
In the rectangle below, a charge is to be placed
at the empty corner to make a net force on the
charge at corner A point along the vertical
direction. What charge (magnitude and sign)
must be placed at the empty corner?
ANSWER: -3.3 µC
+3 µC
4d
C
B
d
A
+3 µC
D
+3 µC
+1µC
Electric field
?
d = 5cm
B
C
A
D
d
a) Find the magnitude & direction -2µC
+1µC
of the electric field at corner C
of the square. ANS: 1.5 x 106 N/C at 45°
b) What charge qC should be placed at corner
C so the net electric field at the center of
the square is zero? ANS: -2µC
c) Determine the magnitude and direction of
the force exerted by three charges at A, B,
and D on the fourth charge at C.
ANS: 3N at 225°
Electric potential
+1 µC
A
d = 5cm
?
B
d
d
C
-2 µC
D
d
+1 µC
E
a) What is the total potential at C? ANS: -1.8 x 105 J/C
b) A fourth charge q is placed at B such that
the net potential at C is zero. What is the
magnitude & sign of q? ANS: +1 µC
c) How much work is done to move q from
ANS: +0.06 J
infinity to B?
Uniform electric field
An electron with speed
–
vo = 1.8 x 106 m/s is traveling
parallel to an electric field of
magnitude E = 7.7 x 103 N/C.
(me = 9.11 x 10-31 kg,
e = 1.6 x 10-19C)
a) How far will it travel
before it stops? ANS: 1.2 mm
b) How much time will
+
elapse before it returns
to its starting point? ANS: 2.6 ns
–
–
–
–
–
–
+
+
+
–
+
+
+
A good challenge
An electron is accelerated horizontally from rest in a
TV picture tube by a potential difference of 25,000 V
(energy per charge). It then passes between two
horizontal plates 6.0cm long and 1.3cm apart that
have a potential difference of 250 V. At what angle θ
will the electron be traveling after it passes between
the plates?
–
–
–
–
–
–
–
–
–
–
v = 9.37 x 107m/s
F = 3.07 x 10-15N
a = 3.38 x 1015m/s2
–
ANSWER: 1.32°
+
+
+
+
+
+
+
t = 6.40 x 10-10s
vx = 9.37 x 107m/s
vy = 2.16 x 106m/s
+
+
θ