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
```

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
```

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
```

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
```

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
```

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
```