# Chapter 3 Electrostatic potential and flux¶

## Example 3.1 Page no 89¶

In :
#Given
q=300*10**-6                   #c
V=6

#Calculation
W=q*V

#Result
print"Work done is ", W*10**3,"*10**-3 J"

Work done is  1.8 *10**-3 J


## Example 3.2 Page no 89¶

In :
#given
Va=-10              #V
W=300                #J
q=3.0                  #C

#Calculation
V=(W/q)+Va

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

The value of V is  90.0 Volts


## Example 3.3 Page no 89¶

In :
#Given
m=9*10**9
q=16*10**-10                #C
r=0.1
r1=0.06
q1=12*10**-10

#Calculation
Vb=m*q/r
Vb1=m*q/r1
V=Vb1-Vb
W=q1*V

#Result
print"Workdone is ", W*10**8,"*10**-8 J"

Workdone is  11.52 *10**-8 J


## Example 3.4 Page no 89¶

In :
#Given
r=3.4*10**-14                  #m
n=47
q=1.6*10**-19              #C
m=9*10**9

#Calculation
V=m*n*q/r

#Result
print"Electric potential at the surface of silver nucleus is ", round(V*10**-6,2),"*10**6 V"

Electric potential at the surface of silver nucleus is  1.99 *10**6 V


## Example 3.5 Page no 90¶

In :
#Given
m=9*10**9
q=4*10**-6

#Calculation
V=2*q*m

#Result
print"Electric potential is ", V*10**-3,"*10**3 V"

Electric potential is  72.0 *10**3 V


## Example 3.9 Page no 91¶

In :
#Given
m=9*10**9
q=250*10**-6
r=0.1

#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  2.25 *10**7 V


## Example 3.10 Page no 96¶

In :
#Given
m=3*10**-16
g=9.8
d=5*10**-3
q=16.0*10**-18

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

#Result
print"Voltage needed to balance an oil drop is ",round(V,2),"V"

Voltage needed to balance an oil drop is  9.19 V


## Example 3.12 Page no 96¶

In :
#Given
q=1.6*10**-19                  #C
V=3000                         #V
r=5*10**-2                     #m
g=9.8

#Calculation
E=V/r
m=q*E/g

#Result
print"The mass of the particle is ", round(m*10**16,1),"*10**-16 Kg"

The mass of the particle is  9.8 *10**-16 Kg


## Example 3.13 Page no 100¶

In :
#Given
m=9*10**-9
q1=3*10**-9
q2=3*10**-9
q3=10**9
r=0.2

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

#Result
print"Workdone is ", W,"J"

Workdone is  2.7e-07 J


## Example 3.14 Page no 100¶

In :
#Given
m=9*10**9
q=1.6*10**-19
r=10**-10

#Calculation
U=m*q**2/r
K=U/2.0

#Result
print"Kinetic energy is ",K,"J"

Kinetic energy is  1.152e-18 J


## Example 3.15 Page no 100¶

In :
#Given
m=9*10**-31
V=10**6
q=1.6*10**-19
a=9*10**9

#Calculation
K=m*V**2
r=a*q**2/K

#Result
print"Distance of the closest approach is ", r,"m"

Distance of the closest approach is  2.56e-10 m


## Example 3.17 Page no 101¶

In :
#Given
r=0.53*10**-10                   #m
q1=1.6*10**-19                   #C
q2=-1.6*10**-19                  #C
a=9*10**9
r1=1.06*10**-10

#Calculation
U=a*q1*q2/r
Ue=U/q1
K=-Ue/2.0
E=Ue+K
U1=(a*q1*q2/r1)/q1

#Result
print"(i) Potential energy of the system is ", round(Ue,1),"eV"
print"(ii) Minimum amount of work required to free the elctrons ia ",round(E,1),"ev"
print"(iii) Potential energyof the system is ",round(E,1) ,"ev and work requiredto free the electrons is ",round(-E,1),"eV"

(i) Potential energy of the system is  -27.2 eV
(ii) Minimum amount of work required to free the elctrons ia  -13.6 ev
(iii) Potential energyof the system is  -13.6 ev and work requiredto free the electrons is  13.6 eV


## Example 3.18 Page no 102¶

In :
#Given
a=9*10**9
q1=7*10**-6                          #C
q2=-2*10**-6
r=0.18
r1=0.09
A=9*10**5

#Calculation
U=a*q1*q2/r
W=0-U
U1=(q1*A/r1)+(q2*A/r1)+U

#Result
print"(a) Electrostatic potential energy is ", round(U,1),"J"
print"(b) Work required to seperate two charges is ",round(W,1),"J"
print"(c) Electrostatic energy is ", U1,"J"

(a) Electrostatic potential energy is  -0.7 J
(b) Work required to seperate two charges is  0.7 J
(c) Electrostatic energy is  49.3 J


## Example 3.20 Page no 103¶

In :
#Given
p=6*10**-6
E=10**6
a=1

#Calculation,
U1=-p*E*a
U2=(p*E*(math.cos(60)*180/3.14))*10**-2
U3=U2-U1

#Result
print"Heat released by substance is ", round(U3,0),"J"

Heat released by substance is  3.0 J


## Example 3.21 Page no 109¶

In :
#Given
q=10**-7
e=8.854*10**-12

#Calculation
a=q/e

#Result
print"Electric flux through the surface of the cube is ", round(a*10**-4,2),"Nm**2C-1"

Electric flux through the surface of the cube is  1.13 Nm**2C-1


## Example 3.22 Page no 109¶

In :
#Given
q=8.85*10**-6
e=8.85*10**-12

#Calculation
a=q/e
b=a/6.0

#Result
print"Electric flux through each face is ", round(b*10**-5,2),"Nm**2C-1"

Electric flux through each face is  1.67 Nm**2C-1


## Example 3.23 Page no 109¶

In :
#Given
E0=2*10**3                    #N/C
S=0.2

#Calculation
a=(3/5.0)*E0*S

#Result
print"Electric flux of the field is ", a,"Nm**2C-1"

Electric flux of the field is  240.0 Nm**2C-1


## Example 3.24 Page no 109¶

In :
#Given
r=0.2
m=9*10**9
b=50

import math
E=250*r
a=E*4*math.pi*r**2
q=b*r**2/m

#Result
print"Charge contained in a sphere is ", round(q*10**10,2)*10**-10,"C"

Charge contained in a sphere is  2.22e-10 C


## Example 3.25 Page no 110¶

In :
#Given
a=0.1                 #m
A=800
e=8.854*10**-12

#Calculation
b=A*a**2.5*(math.sqrt(2)-1)
q=e*b

#Result
print"(a) The flux through the cube is ", round(b,2),"Nm**2C-1"
print"The charge within the cube is ",round(q*10**12,2)*10**-12,"C"

(a) The flux through the cube is  1.05 Nm**2C-1
The charge within the cube is  9.28e-12 C


## Example 3.26 Page no 111¶

In :
#Given
E=200
a=0.05
e=8.854*10**-12
d=3.14

#Calculation
import math
b=E*math.pi*a**2
c=2*b
q=e*d

#Result
print"(a) Net outward flux through each flat face is ", round(b,2),"Nm**2C-1"
print"(b) Flux through the side of cylinder is zero "
print"(c) Net outward flux through the cylinder is ", round(c,2),"Nm**2C-1"
print"(d) The net charge in the cylinder is ",round(q*10**11,2)*10**-11,"C"

(a) Net outward flux through each flat face is  1.57 Nm**2C-1
(b) Flux through the side of cylinder is zero
(c) Net outward flux through the cylinder is  3.14 Nm**2C-1
(d) The net charge in the cylinder is  2.78e-11 C


## Example 3.28 Page no 114¶

In :
#Given
q=5.8*10**-6               #C
r=8*10**-2                 #m
e=8.854*10**-12
l=3.0

#Calculation
import math
E=q/(2*math.pi*e*r*l)

#Result
print"Electric field is ", round(E*10**-5,1),"*10**5 N/C"

Electric field is  4.3 *10**5 N/C


## Example 3.29 Page no 114¶

In :
#Given
E=9*10**4                   #N/C
r=2*10**-2                  #m
m=9*10**9

#Calculation
a=r*E/(2.0*m)
print"Linear charge density is ", a,"Cm-1"

Linear charge density is  1e-07 Cm-1


## Example 3.30 Page no 115¶

In :
#Given
q=10*10**-6                   #C
r=0.1                         #m
a=8.85*10**-12

#Calculation
import math
E=q/(4.0*math.pi*a*r**2)

#Result
print"(i) Electric field intensity at a point 10 cm from the centre", round(E*10**-6,0),"*10**6 N/C"
print"(ii) Since the point is lying inside the shell, electric intensity at this point is zero"

(i) Electric field intensity at a point 10 cm from the centre 9.0 *10**6 N/C
(ii) Since the point is lying inside the shell, electric intensity at this point is zero


## Example 3.31 Page no 118¶

In :
#Given
Z=79
e=1.6*10**-19
e0=8.854*10**-12
R=6.2*10**-15

#Calculation
import math
q=Z*e
E=q/(4.0*math.pi*e0*R**2)
b=E/4.0

#Result
print"(i) The magnitude of the electric field at the surface of nucleus is ", round(E*10**-21,0)*10**21,"N/C"
print"(ii) The magnitude of the electric field at a distance 2R from the centre of the nucleus is ",round(b*10**-21,2),"*10**21 N/C"

(i) The magnitude of the electric field at the surface of nucleus is  3e+21 N/C
(ii) The magnitude of the electric field at a distance 2R from the centre of the nucleus is  0.74 *10**21 N/C


## Example 3.32 Page no 119¶

In :
#Given
e=8.854*10**-12
A=0.5
F=1.8*10**-12                   #N
E=1.6*10**-19

#Calculation
q=(2*e*A**2*F)/E

#Result
print"Total charge on the sheet is ", round(q*10**6,0),"micro C"

Total charge on the sheet is  50.0 micro C


## Example 3.33 Page no 119¶

In :
#Given
a=5*10**-6
e=8.854*10**-12
r=0.1

#Calculation
import math
b=-(a*math.pi*r**2*(math.cos(60)*180/3.14))/(2*e)

#Result
print"Electric flux through a circular area is ", round(b*10**-5,2),"*10**3 Nm**2C-1"

Electric flux through a circular area is  4.84 *10**3 Nm**2C-1