Chapter 3 Electric potential

Example 3.1 Page no 65

In [4]:
#Given
q=4.0                               #c
Va=-10                            #V
Wab=100                           #joule

#Calculation
V=(Wab/q)+Va

#Result
print"The value of V is", V,"Volts"

The value of V is 15.0 Volts

Example 3.2 Page no 65

In [10]:
#Given
Z=79
e=1.6*10**-19                      #C
r=6.6*10**-15                      #m
m=9*10**9

#Calculation
q=Z*e
V=(m*q)/r

#Result
print"Electric potential at the surface of a gold nucleus is", round(V*10**-7,3),"*10**7 V"

Electric potential at the surface of a gold nucleus is 1.724 *10**7 V

Example 3.3 Page no 65

In [17]:
#Given
q=4*10**-7                         #C
r=0.09                              #m
m=9*10**9
a=2*10**-9

#Calculation
Vp=(m*q)/r
W=a*Vp

#Result
print"(a) Potential at point P is", Vp*10**-4,"*10**4 V"
print"(b) Work done is", W,"J"

(a) Potential at point P is 4.0 *10**4 V
(b) Work done is 8e-05 J

Example 3.4 Page no 66

In [29]:
#Given
q=20*10**-6                              #C
r1=0.1                                  #m
r2=0.05                                   #m
m=9*10**9
e=1.6*10**-19

#Calculation
Va=(m*q)/r1
Vb=(m*q)/r2
Wab=-(Va-Vb)*e

#Result
print"Potential at point A is", Va*10**-6,"*10**6 V"
print"Potential at point b is", Vb*10**-6,"*10**6 V"
print"Work done is",Wab,"J"

Potential at point A is 1.8 *10**6 V
Potential at point b is 3.6 *10**6 V
Work done is 2.88e-13 J

Example 3.5 Page no 66

In [35]:
#Given
m=9.0*10**9
r=0.5
a=16

#Calculation
q=(a*r)/m

#Result
print"Magnitude of charge is", round(q*10**10,2)*10**-10,"C"

Magnitude of charge is 8.89e-10 C

Example 3.6 Page no 66

In [41]:
#Given
q1=2*10**-6                          #c
q2=-2*10**-6
q3=-3*10**-6
q4=6*10**-6
r=1.0                                  #m
m=9*10**9

#Calculation
V=m*(q1+q2+q3+q4)/r

#Result
print"Potential at the centre of the square is",V*10**-4,"*10**4 V"

Potential at the centre of the square is 2.7 *10**4 V

Example 3.7 Page no 66

In [45]:
#Given
m=9*10**9
q=5*10**-6
r=0.1

#Calculation
V=(6*m*q)/r

#Result
print"Potential at the centre of the hexagon is", V*10**-6,"*10**6 V"

Potential at the centre of the hexagon is 2.7 *10**6 V

Example 3.9 Page no 67

In [50]:
#Given
r=0.1                            #m
q=200*10**-6                     #C
m=9*10**9

#Calculation
V=(m*q)/r

#Result
print"Electric potential at the centre is", V*10**-7,"*10**7 V"

Electric potential at the centre is 1.8 *10**7 V

Example 3.11 Page no 67

In [62]:
#Given
a=0.2                         #m
m=9*10**9
q1=2*10**-9
q2=4*10**-9
q3=8*10**-9

#Calculation
import math
V=m*((q1/a)+(q2/(a*math.sqrt(2)))+(q3/a))
b=0.1*math.sqrt(2)
V0=m*((q1/b)+(q2/b)+(q3/b))
W=q*(V0-V)

#Result
print"Required work is",round(W*10**2,2)*10**-7,"J"

Required work is 6.27e-07 J

Example 3.12 Page no 68

In [66]:
#Given
V=60
r=0.05

#Calculation
E=-V/r

#Result
print"Electric field intensity between the two plate is", E,"V/m"

Electric field intensity between the two plate is -1200.0 V/m

Example 3.13 Page no 68

In [71]:
#Given
e=1.6*10**-19
m=3*10**-16                             #Kg
r=5*10**-3                              #m
g=9.8

#Calculation
q=10*e
V=(m*g*r)/q

#Result
print round(V,2),"V"

9.19 V

Example 3.14 Page no 68

In [78]:
#Given
q1=7*10**-6                     #c
q2=-2*10**-6                    #C
r12=0.18                        #m
m=9*10**9

#Calculation
U=m*((q1*q2)/r12)
W=0-U

#Result
print"(a) Electrostatic potential energy of a system is",U,"J"
print"(b) work required is", W,"J"

(a) Electrostatic potential energy of a system is -0.7 J
(b) work required is 0.7 J

Example 3.15 Page no 68

In [91]:
#Given
m=9*10**9
e=1.6*10**-19                             #C
r=5.3*10**-11                             #m

#Calculation
V=(m*e)/r
U=((m*e*-e)/r)/e
q=2*e
V1=(m*q)/r

#Result
print"(a) Electric potential is", round(V,2),"V"
print"(b) Electric potential energy is",round(V1,2),"V"

(a) Electric potential is 27.17 V
(b) Electric potential energy is 54.34 V

Example 3.16 Page no 69

In [109]:
#Given
r12=1.5*10**-10                    #m
r13=10.0**-10
q1=1.6*10**-19                        #C
q3=-1.6*10**-19
m=9*10**9

#Calculation
U=m*(((q1**2)/r12)+((q1*q3)/r13)+((q1*q3)/r13))/q1

#Result
print"Potential energy is",U,"ev"

Potential energy is -19.2 ev

Example 3.17 Page no 69

In [126]:
#Given
E=10**5                           #N/C
q=10**-6                      #C
a=0.02                           #m
A=1

#Calculation'
import math
p=q*a
t=p*E*A
W=p*E*(math.cos(0*3.14/180.0)-math.cos(180*3.14/180.0))

#Result
print"(i) Torque acting on dipole is",t,"N m"
print"(ii) Work done is", round(W,3),"J"

(i) Torque acting on dipole is 0.002 N m
(ii) Work done is 0.004 J

Example 3.18 Page no 69

In [129]:
#Given
m=9*10**9
q=8*10**-3
q0=-2*10**-9
ra=0.03
rb=0.04

#Calculation
W=(m*q*q0*(1/rb-1/ra))

#Result
print"Work done is", W,"J"

Work done is 1.2 J

Example 3.19 Page no 69

In [160]:
#Given
qa=1.5*10**-6                     #C
qb=2.5*10**-6 
a=0.15                                 #m
m=9*10**9
a1=0.18

#Calculation
import math
V=((qa+qb)/a)*m
Ea=m*qa/a**2
Eb=m*qb/a**2
E=Eb-Ea
V1=((qa+qb)/a1)*m
Ea1=m*qa/a1**2
Ea2=m*qb/a1**2
A=math.atan(15/10.0)*180/3.14
A1=2*A
A2=math.cos(A1*3.14/180.0)
E3=math.sqrt(Ea1**2+Ea2**2+(2*Ea1*Ea2*A2))
tb=Ea1*math.sin(A1*3.14/180.0)/(Ea2+Ea1*math.cos(A1*3.14/180.0))
B=math.atan(tb)*180/3.14
y=90-A
Y=y+B

#Result
print "(a) Net electric field is",E*10**-5,"*10**5 N/C"
print"(b) Angle is" ,round(Y,1),"degree"

(a) Net electric field is 4.0 *10**5 N/C
(b) Angle is 69.4 degree

Example 3.23 Page no 71

In [181]:
#Given
q1=-1.6*10**-19                  #C
q2=1.6*10**19
r12=0.53*10**-10                  #m
m=9*10**9
r11=1.06*10**-10

#Calculation
U=((m*q1*q2)/r12)/q2
K=U/2.0
E=-(U-K)
U1=((m*q1*q2)/r11)/q2
E1=U-U1

#Result
print "(a) Potential energy is",round(U,2),"ev"
print"(b) Minimum work required is",round(E,3) ,"ev"
print"(c) Potential energy is",round(E1,3),"ev"

(a) Potential energy is -27.17 ev
(b) Minimum work required is 13.585 ev
(c) Potential energy is -13.585 ev

Example 3.24 Page no 72

In [188]:
#Given
E=10**6                         #V/m
p=10**-29                       #cm
a=6.023*10**23

#Calculation
import math
Np=a*p
U1=Np*E
U2=Np*E*math.cos(60*3.14/180.0)
U=U1-U2

#Result
print"Heat released by the substance is", round(U,2),"J"

Heat released by the substance is 3.01 J