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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 10: Electrostatics"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.1, Page 507"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"m = 4e-013; # Mass of the particle, kg\n",
"q = 2.4e-019; # Charge on particle, C\n",
"d = 2e-002; # Distance between the two horizontally charged plates, m\n",
"g = 9.8; #`Acceleration due to gravity, m/sec-square\n",
"\n",
"#Calculations\n",
"E = (m*g)/q ; # Electric field strength, N/C\n",
"V = E*d; # Potential difference between the two charged horizontal plates, V\n",
"\n",
"#Result\n",
"print \"The potential difference between the two horizontally charged plates = %3.1e V\"%V\n",
"#Incorrect answer in textbook"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The potential difference between the two horizontally charged plates = 3.3e+05 V\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.2, Page 507"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import *\n",
"\n",
"#Variable Declaration\n",
"q1 = 1e-009; # Charge at first corner, C\n",
"q2 = 2e-009; # Charge at second corner, C\n",
"q3 = 3e-009; # Charge at third corner, C\n",
"d = 1.; # Side of the equilateral triangle, m\n",
"theta = 30; # Angle at which line joining the observation point to the source charge makes with the side, degrees\n",
"\n",
"#Calculations\n",
"r = (d/2)/cos(theta*pi/180); # Distance of observation point from the charges, m\n",
"#since,1/4*%pi*%eps = 9e+009;\n",
"V = (q1+q2+q3)*(9e+009)/r; # Elecric potential, V\n",
"\n",
"#Result\n",
"print \"The electric potential at the point equidistant from the three corners of the triangle = %4.1f V\"%V\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electric potential at the point equidistant from the three corners of the triangle = 93.5 V\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.3, Page 508"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import *\n",
"\n",
"#Variable Declaration\n",
"q = 2e-008; \n",
"q1 = q; # Charge at first corner, C\n",
"q2 = -2*q; # Charge at second corner, C\n",
"q3 = 3*q; # Charge at third corner, C\n",
"q4 = 2*q; # Charge at fourth corner, C\n",
"d = 1; # Side of the square, m\n",
"\n",
"#Calculations\n",
"r = round(d*sin(45*pi/180),2); # Distance of centre of the square from each corner, m\n",
"V = (q1+q2+q3+q4)*(9e+009)/r; # Elecric potential at the centre of the square, V \n",
"\n",
"#Result\n",
"print \"The electric potential at the centre of the square = %4d V\"%V\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electric potential at the centre of the square = 1014 V\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.6, Page 511"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"V = 60; # Electric potential of smaller drop, volt\n",
"r = 1; # For simplcity assume radius of each small drop to be unity, unit\n",
"q = 1; # For simplicity assume charge on smaller drop to be unity, C\n",
"k = 1; # For simplicity assume Coulomb's constant to be unity, unit\n",
"\n",
"#Calculations\n",
"R = 2**(1./3)*r; # Radius of bigger drop, unit\n",
"Q = 2*q; # Charge on bigger drop, C\n",
"V_prime = k*Q/R*V; # Electric potential of bigger drop, volt\n",
"\n",
"#Result\n",
"print \"The electric potential of new drop = %4.1f V\"%V_prime\n",
"#Incorrect answer in the textbook"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electric potential of new drop = 95.2 V\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.7, Page 512"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"m = 9.1e-031; # Mass of the electron, kg\n",
"e = 1.6e-019; # Charge on an electron, C\n",
"g = 9.8; # Acceleration due to gravity, m/sec-square\n",
"\n",
"#Calculations\n",
"# Electric force, F = e*E, where F = m*g or e*E = m*g\n",
"E = m*g/e; # Electric field which would balance the weight of an electron placed in it, N/C\n",
"\n",
"#Result\n",
"print \"The required electric field strength = %3.1e N/C\"%E\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required electric field strength = 5.6e-11 N/C\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.8, Page 512"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable Declaration\n",
"q1 = 8e-007; # First Charge, C\n",
"q2 = -8e-007; # Second Charge, C\n",
"r = 15e-002; # Distance between the two charges, m\n",
"k = 9e+009; # Coulomb's constant, N-metre-square/coulomb-square\n",
"\n",
"#Calculations&#Results\n",
"E1 = k*q1/r**2; # Electric field strength due to charge 8e-007 C\n",
"print \"The electric field strength at midpoint = %3.1e N/C\"%E1\n",
"E2 = abs(k*q2/r**2); # Electric field strength -8e-007 C \n",
"print \"The electric field strength at midpoint = %3.1e N/C\"%E2\n",
"# Total electric field strength at the mid-point is\n",
"E = E1+E2; # Net electric field at mid point, N/C\n",
"print \"The net electric field strength at midpoint = %3.1e N/C\"%E\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electric field strength at midpoint = 3.2e+05 N/C\n",
"The electric field strength at midpoint = 3.2e+05 N/C\n",
"The net electric field strength at midpoint = 6.4e+05 N/C\n"
]
}
],
"prompt_number": 21
}
],
"metadata": {}
}
]
}
|
