Chapter 10: Electrostatics

Example 10.1, Page 507

In [9]:
#Variable Declaration
m = 4e-013;  # Mass of the particle, kg
q = 2.4e-019;  # Charge on particle, C
d = 2e-002;  # Distance between the two horizontally charged plates, m
g = 9.8;  #`Acceleration due to gravity, m/sec-square

#Calculations
E = (m*g)/q ;  # Electric field strength, N/C
V = E*d;     # Potential difference between the two charged horizontal plates, V

#Result
print "The potential difference between the two horizontally charged plates = %3.1e V"%V
#Incorrect answer in textbook

The potential difference between the two horizontally charged plates = 3.3e+05 V

Example 10.2, Page 507

In [10]:
from math import *

#Variable Declaration
q1 = 1e-009;    # Charge at first corner, C
q2 = 2e-009;    # Charge at second corner, C
q3 = 3e-009;    # Charge at third corner, C
d = 1.;  # Side of the equilateral triangle, m
theta = 30;  # Angle at which line joining the observation point to the source charge makes with the side, degrees

#Calculations
r = (d/2)/cos(theta*pi/180);  # Distance of observation point from the charges, m
#since,1/4*%pi*%eps = 9e+009;
V = (q1+q2+q3)*(9e+009)/r;  # Elecric potential, V

#Result
print "The electric potential at the point equidistant from the three corners of the triangle = %4.1f V"%V

The electric potential at the point equidistant from the three corners of the triangle = 93.5 V

Example 10.3, Page 508

In [17]:
from math import *

#Variable Declaration
q = 2e-008;    
q1 = q;        # Charge at first corner, C
q2 = -2*q;     # Charge at second corner, C
q3 = 3*q;      # Charge at third corner, C
q4 = 2*q;      # Charge at fourth corner, C
d = 1;      # Side of the square, m

#Calculations
r = round(d*sin(45*pi/180),2);  # Distance of centre of the square from each corner, m
V = (q1+q2+q3+q4)*(9e+009)/r;  # Elecric potential at the centre of the square, V 

#Result
print "The electric potential at the centre of the square = %4d V"%V

The electric potential at the centre of the square = 1014 V

Example 10.6, Page 511

In [19]:
#Variable Declaration
V = 60;  # Electric potential of smaller drop, volt
r = 1;  # For simplcity assume radius of each small drop to be unity, unit
q = 1;  # For simplicity assume charge on smaller drop to be unity, C
k = 1;  # For simplicity assume Coulomb's constant to be unity, unit

#Calculations
R = 2**(1./3)*r;    # Radius of bigger drop, unit
Q = 2*q;  # Charge on bigger drop, C
V_prime = k*Q/R*V;  # Electric potential of bigger drop, volt

#Result
print "The electric potential of new drop = %4.1f V"%V_prime
#Incorrect answer in the textbook

The electric potential of new drop = 95.2 V

Example 10.7, Page 512

In [20]:
#Variable Declaration
m = 9.1e-031;  # Mass of the electron, kg
e = 1.6e-019;  # Charge on an electron, C
g = 9.8;       # Acceleration due to gravity, m/sec-square

#Calculations
# Electric force, F = e*E, where F = m*g or e*E = m*g
E = m*g/e;    # Electric field which would balance the weight of an electron placed in it, N/C

#Result
print "The required electric field strength = %3.1e N/C"%E

The required electric field strength = 5.6e-11 N/C

Example 10.8, Page 512

In [21]:
#Variable Declaration
q1 = 8e-007;  # First Charge, C
q2 = -8e-007;  # Second Charge, C
r = 15e-002;  # Distance between the two charges, m
k = 9e+009; # Coulomb's constant, N-metre-square/coulomb-square

#Calculations&#Results
E1 = k*q1/r**2;  # Electric field strength due to charge 8e-007 C
print "The electric field strength at midpoint = %3.1e N/C"%E1
E2 = abs(k*q2/r**2);  # Electric field strength -8e-007 C 
print "The electric field strength at midpoint = %3.1e N/C"%E2
# Total electric field strength at the mid-point is
E = E1+E2;      # Net electric field at mid point, N/C
print "The net electric field strength at midpoint = %3.1e N/C"%E

The electric field strength at midpoint = 3.2e+05 N/C
The electric field strength at midpoint = 3.2e+05 N/C
The net electric field strength at midpoint = 6.4e+05 N/C