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Phy213: General Physics III
Chapter 22 Worksheet
6/30/2017
1
Electric Fields:
1. Consider a fixed point charge of +2.0 C (q1).
q1 = +2.0 C
P
.
d = 0.1 m
a.
What is the magnitude and direction of the electric field at a point P, a distance of 0.1 m?
Ans.
b.
 1  q1 ˆ
6 N ˆ
E= 
 2 i= 1.8x10 C i
 4 o  d


A 2nd charge (q2 = -2.0 C) is placed at point P. What is the magnitude and direction of the
electric force exerted on q3?
Ans.
 1  q1q2 ˆ
ˆ
FE = 
 2 i = -q2E = - 3.6 N i {an attractive force}
4

d
o 

2. Three protons are present in the nucleus of a lithium atom, forming an equilateral triangle.
The distance, r, between each pair of protons is 1.5 x 10-15 m.
Lithium Nucleus
r
+
+
+
a. What is the magnitude of the electric field vector due to the lithium nucleus at point P, a
distance of 2.5x10-10m from the center of the nucleus?
Ans.
Since d>>r: E
 1  3qp
10 N

 2 = 6.9x10 C {radial outward from nucleus}
 4 o  d
b. What is the magnitude and direction of the total electric force exerted on an electron
positioned at point P.
Ans.
FE = qe E = e E = 1.1x10-8 N {inward towatd nucleus}
Phy213: General Physics III
Chapter 22 Worksheet
6/30/2017
2
3. Consider an electric dipole, q = 0.1 C, separated by a distance (d) of 5x10-6 m.
+q
+
-q
-
P
.
r+

E-
r-
E+
a.
Sketch the electric field lines for the electric field vector due to the dipole.
b
What is the magnitude and direction of the electric field vector due to the dipole at point P, a
distance (r) of 0.5 m to the right of the dipole?
 q 
 q 
E = E++ E- =
cos ˆi - sin ˆj + 
- cos ˆi - sin ˆj
2 
2 
4

r
4

r
o + 
o - 



Ans.
c.
Ans.


 - 2q 
 2d 
ˆj =  -q  
E  
sin


2 
2 
2
 4 or 
 2 or   r2+  2d 




 ˆj =  -p  ˆj = - 3.6 10-2

3 

 4 or 

N
C
ˆj
What is the magnitude and direction of the electric field due to the dipole at a distance of
0.5 m directly above the dipole?
E
 1 p ˆ
-2 N ˆ

 3 j = 7.2x10 C j
 2 o  r
4. Consider a uniform line of charge with a total charge of 5 pC and a total length of 0.5 m.
dE
P
.
dE
y
dq
dq
+++++++++++++++++++++++++++++++++++
++++
x
a.
Ans.
What is the magnitude of the linear charge density?
 =
5 pC
= 1x10-11
0.5 m
C
m
b.
Sketch the field lines for the line of charge in the diagram above.
c.
Derive an equation for the electric field vector at point P, located 0.3 m above the line and
centered between the ends.
Ans.
Since all but the vertical component of E cancels out:

 1  dq
 1   dx  
y
ˆ
dEy = d E cos = 
 2 cos j = 
 2
2 
 4 o  r
 4 o   x + y   x2 + y2


Solving for E:



1
2

 ˆj


Phy213: General Physics III
Chapter 22 Worksheet
6/30/2017
3


 y  L 
dx
 ˆj
E =  dE = 
3
 2

2
2
 4 o  0  x + y


L
ˆj {in general, radially outward from line}
E=
1
2
2 2
4 o y L + y


d.
Ans.
e.
Ans.
f.
Ans.


What is the magnitude of the electric field vector at point P?
E = 0.44 NC {directed away from the line of charge}
Determine the electric force vector exerted on an electron placed at point P.
FE= qeE = eE = -7.0x10-20 N ˆj{inward toward the line}
What is the acceleration of the electron located at point P?
a=
FE
= -7.7x1010
me
m
s2
ˆj {inward toward the line}
5. Consider a charged ring, with total charge of 2.4 mC and radius of 0.3 m.
-q
a.
Ans.
b.
Ans.
P
.
What is the magnitude of the surface charge density?
 =
2.4x10-3C
 0.3 m
2
= 8.5x10-3 mC2
Derive an equation for the electric field vector at point P, located 1.5 m to the right of the
center of the disc.
Since all but the radial component of E cancels out, establish dEcos in terms of dq:

 1  dq
 1    rdr  
x
ˆ
d E cos =
 2 cos i=
 2
2 
 4 o  d
 4 o   x + r   x2 + r2



Solving for E:


1
2


 ˆi= 1    xrdr
  4 o   2
2

 x +r


3
2

 ˆi


Phy213: General Physics III
Chapter 22 Worksheet
6/30/2017
4

 x  R
rdr
E =  dE = 
 2
 4 o  0  x + r 2



3
2

 ˆi


R
 x 
E= 

 4 o 
-1
x
2
+R

1
ˆi =   x   1 

1
 4 o   x
x2 + R 2
2 2



0

1
2

 ˆi {away from ring}


Phy213: General Physics III
Chapter 22 Worksheet
d.
6/30/2017
5
What is the magnitude and direction of the electric field vector at point P?
Ans.
(8.5x10-3 mC2 )(1.5m) 
1
1
E=
 (1.5m)
4 o
(1.5m)2 + (0.3m)2

E = 1.5x106 NC ˆi {away from ring}


1
2

 ˆi


5. The plates of a parallel plate capacitor have a uniform charge of +1.0 C and -1.0 C,
respectively. Each plate is a square with sides of length of 0.5 m.
a. What is the magnitude of the surface charge density on each plate?
Ans.
 =
1.0C
0.5 m
2
= 4.0
C
m2
b. What is the magnitude of the electric field vector inside the plates of the capacitor?
Ans.
Treat each plate as a charged flat disc of infinite R, i.e. x << R, then the E field due to
each plate respectively is then:
  ˆ
Eplate = 
 i {both fields will point toward the right}
2

 o
The total electric field between the plates is therefore:
 
Ebetween plates = Eplate 1 + Eplate 2 =   ˆi = 4.5x1011 NC ˆi
 o 
c. What is the electric force exerted on an electron placed inside the plates of the capacitor?
Ans.
FE= qeE = eE = -7.2x10-8 N ˆj {to the left}
d. What is the acceleration of the electron while it is in between the capacitor plates?
Ans.
e.
Ans.
a=
FE
= -7.9x1022
me
m
s2
ˆj {to the left}
For a 0.2 m plate separation, calculate the final speed of an electron (initially at rest), that
is placed at the negative charged plate and just reaches the positive plate.
v=
v2 = -


2ad ˆi = - 2 -7.9x1022 sm2 -0.2m ˆi
v = -1.8x1011 ˆi {a bit unrealistic eh??}
Phy213: General Physics III
Chapter 22 Worksheet
6/30/2017
6
Dipole Moment of Water:
6. Water is a polar molecule that consists of 2 H atoms attached to a central oxygen atom.
The H atoms are oriented at an angle of approximately 105o. Each of the H-O bonds has a
dipole moment ( pOH ) associated with it and together the 2 dipole moments have a resulting
net dipole moment ( pH2O ) of 6.2x10-30 C.m. The effective separation distance between the
respective positive (+) and negative (-) charges in each O-H bond is 3.9x10-12 m.
a.
What is the magnitude of the effective dipole moment of
the individual O-H bond for water?
Ans.

pOH = pOH1 = pOH2 =
b.
= 5.1x10 C  m
H
=105O
pOH = qd =  d= 5.1x10-30C  m   = 1.3x10-18C
Calculate the torque on a water molecule due to the electric field of a charged sphere
(qsphere = -1.5x10-6 C) at a separation distance of 0.05 m, where the direction of the
electric field is oriented at an angle of 30o to pH2O .
Ans.
 1  qsphere
6 N
E = 
 2 = 5.4x10 C
 4 o  r


 E = p  E = pEsin30o = 1.7x10-23 N  m
d.

Determine the magnitude of the charge () for the individual O-H dipole.
Ans.
c.
-10
2cos(52.5o )

H
pH2O = pOH1 + pOH2 = 2qdcos(52.5o )jˆ =
pH2O
 O
O
What is the potential energy of the water molecule in this electric field?
Ans.
U = p  E = pEcos30o = 2.9x10-23 J