Chapter 15:Electric Forces and Fields

Ex15.1:pg-719

In [1]:
  import math   #Example15_1
 
  #To find the value of q and how many electrons must be removed and fraction of atoms lost
dist=2       #Units in meters
f=0.0294     #Units in N
s=9*10**9   #Units in N meter**2/C**2
q=math.sqrt((dist**2*f)/s)    #Units in C
print "The value of q is=",round(q,8)," C\n"
charge=3.61*10**-6   #Units in C
c_elec=1.6*10**-19       #Units in C
n=charge/c_elec     #Units in number
print "Number of electrons to be removed is="
print n
f1=3*10.0**22     #Units in number
fraction=n/f1     #Units of number
print "Fraction of atoms lost is="
print fraction

The value of q is= 3.61e-06  C

Number of electrons to be removed is=
2.25625e+13
Fraction of atoms lost is=
7.52083333333e-10

Ex15.2:pg-721

In [2]:
  import math   #Example15_2
 
  
  #To find the force on the center charge
k=9*10**9      #Units in N meter**2/C**2
q1=4*10.0**-6      #Units in C
q2=5*10.0**-6     #Units in C
r1=2           #Units in meters
r2=4       #Units in meters
q3=6*10.0**-6       #Units in C
f1=(k*q1*q2)/r1**2         #Units in N
f2=(k*q2*q3)/r2**2         #Units in N
f=f1-f2         #Units in C
print "The force on the center charge is=",round(f,5)," N"

The force on the center charge is= 0.02813  N

Ex15.3:pg-722

In [3]:
  import math   #Example15_3
 
  
  #To find the resultant force
f1=6   #Units in N
f2=18       #Units in N
f=math.sqrt(f1**2+f2**2)   #Units in N
theta=math.atan(f2/f1)*180/math.pi    #Units in degrees
print "The resultant force is f=",round(f)," N \nThe resultant angle is theta=",round(theta,1)," degrees"
  #In text book answer printed wrong as f=19 N correct answer is f=18N 

The resultant force is f= 19.0  N 
The resultant angle is theta= 71.6  degrees

Ex15.4:pg-724

In [4]:
  import math   #Example15_4
 

  #To find the resultant force on 20 micro C
f1=2   #Units in N
f2=1.8       #Units in N
theta=37.0         #Units in degrees
f2x=f2*math.cos(theta*math.pi/180)        #Units in N
f2y=f2*math.sin(theta*math.pi/180)        #Units in N
fy=f1+f2y       #Units in N
f=math.sqrt(fy**2+f2x**2)   #Units in N
theta=math.atan(fy/f2x)*180/math.pi    #Units in degrees
print "The resultant force is f=",round(f,1)," N \n The resultant angle is theta=",round(theta,1)," degrees"

The resultant force is f= 3.4  N 
 The resultant angle is theta= 65.0  degrees

Ex15.6:pg-726

In [5]:
  import math   #Example15_6
 
  
  #To find the magnitude of E
k=9*10**9      #Units in N meter**2/C**2
q=3.6*10**-6        #Units in C
theta=37            #Units in degrees
r=10*math.sin(theta*math.pi/180)*10**-2      #Units in meters  
e1=(k*q)/r**2       #Units in N/C
q2=5*10**-6        #Units in C
theta=37            #Units in degrees
r1=10*10**-2   #Units in meters  
e2=(k*q2)/r1**2       #Units in N/C
e1y=e1           #Units in N/C
e2x=e2*math.cos(theta*math.pi/180)      #Units in N/C
e2y=-e2*math.sin(theta*math.pi/180)   #Units in N/C
ex=e2x         #Units in N/C
ey=e1y+e2y          #Units in N/C
e=math.sqrt(ex**2+ey**2)           #Units in N/C
print "The magnitude of E is="
print round(e,2)
print "N/C"
  #In text book the answer isprinted wrong as E=7.26*10**6 N/C but the correct answer is E=7198876.9 N/C

The magnitude of E is=
7198876.9
N/C

Ex15.7:pg-726

In [6]:
  import math   #Example15_7
 
  
  #To find out how much charge occurs
e=3*10.0**6      #Units in N/C
r=0.050      #Units in meters
k=9*10.0**9   #Units in N meter**2/C**2
q=(e*r**2)/k       #Units in C
print "The charge occured is q=",round(q,9)," C"

The charge occured is q= 8.33e-07  C

Ex15.8:pg-727

In [7]:
  import math   #Example15_8
 
  
  #To show using lines of force that a charge suspended with in cavity induces an equal and opposite charge on surface
print "Lines of force come out of positive charge q suspended in cavity.\nCavity \nsurface must possess a negative charge since lines of force go and terminate on q.\nTherefore a charge +q must exist on outer portions."

Lines of force come out of positive charge q suspended in cavity.
Cavity 
surface must possess a negative charge since lines of force go and terminate on q.
Therefore a charge +q must exist on outer portions.

Ex15.9:pg-728

In [8]:
  import math   #Example15_9
  
  #To find the speed just before the field strikes
e=6000     #Units in N/C
q=1.6*10**-19    #Units in C
f=e*q           #Units in N
m=1.67*10**-27          #Units in Kg
a=f/m        #Units in meters/sec**2
vo=0       #Units in meters/sec
x=2*10**-3        #Units in meters
v=math.sqrt(vo**2+(2*a*x))          #Units in meters/sec
print "The field goes by a speed of ",round(v)," meters/sec"
  #In text book answer printed wrong as v=48000 meters/sec the correct answer is v=47952 meters/sec 

The field goes by a speed of  47952.0  meters/sec