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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 15:Electric Forces and Fields"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex15.1:pg-719"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The value of q is= 3.61e-06 C\n",
"\n",
"Number of electrons to be removed is=\n",
"2.25625e+13\n",
"Fraction of atoms lost is=\n",
"7.52083333333e-10\n"
]
}
],
"source": [
" import math #Example15_1\n",
" \n",
" #To find the value of q and how many electrons must be removed and fraction of atoms lost\n",
"dist=2 #Units in meters\n",
"f=0.0294 #Units in N\n",
"s=9*10**9 #Units in N meter**2/C**2\n",
"q=math.sqrt((dist**2*f)/s) #Units in C\n",
"print \"The value of q is=\",round(q,8),\" C\\n\"\n",
"charge=3.61*10**-6 #Units in C\n",
"c_elec=1.6*10**-19 #Units in C\n",
"n=charge/c_elec #Units in number\n",
"print \"Number of electrons to be removed is=\"\n",
"print n\n",
"f1=3*10.0**22 #Units in number\n",
"fraction=n/f1 #Units of number\n",
"print \"Fraction of atoms lost is=\"\n",
"print fraction\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex15.2:pg-721"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The force on the center charge is= 0.02813 N\n"
]
}
],
"source": [
" import math #Example15_2\n",
" \n",
" \n",
" #To find the force on the center charge\n",
"k=9*10**9 #Units in N meter**2/C**2\n",
"q1=4*10.0**-6 #Units in C\n",
"q2=5*10.0**-6 #Units in C\n",
"r1=2 #Units in meters\n",
"r2=4 #Units in meters\n",
"q3=6*10.0**-6 #Units in C\n",
"f1=(k*q1*q2)/r1**2 #Units in N\n",
"f2=(k*q2*q3)/r2**2 #Units in N\n",
"f=f1-f2 #Units in C\n",
"print \"The force on the center charge is=\",round(f,5),\" N\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex15.3:pg-722"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The resultant force is f= 19.0 N \n",
"The resultant angle is theta= 71.6 degrees\n"
]
}
],
"source": [
" import math #Example15_3\n",
" \n",
" \n",
" #To find the resultant force\n",
"f1=6 #Units in N\n",
"f2=18 #Units in N\n",
"f=math.sqrt(f1**2+f2**2) #Units in N\n",
"theta=math.atan(f2/f1)*180/math.pi #Units in degrees\n",
"print \"The resultant force is f=\",round(f),\" N \\nThe resultant angle is theta=\",round(theta,1),\" degrees\"\n",
" #In text book answer printed wrong as f=19 N correct answer is f=18N \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex15.4:pg-724"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The resultant force is f= 3.4 N \n",
" The resultant angle is theta= 65.0 degrees\n"
]
}
],
"source": [
" import math #Example15_4\n",
" \n",
"\n",
" #To find the resultant force on 20 micro C\n",
"f1=2 #Units in N\n",
"f2=1.8 #Units in N\n",
"theta=37.0 #Units in degrees\n",
"f2x=f2*math.cos(theta*math.pi/180) #Units in N\n",
"f2y=f2*math.sin(theta*math.pi/180) #Units in N\n",
"fy=f1+f2y #Units in N\n",
"f=math.sqrt(fy**2+f2x**2) #Units in N\n",
"theta=math.atan(fy/f2x)*180/math.pi #Units in degrees\n",
"print \"The resultant force is f=\",round(f,1),\" N \\n The resultant angle is theta=\",round(theta,1),\" degrees\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex15.6:pg-726 "
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The magnitude of E is=\n",
"7198876.9\n",
"N/C\n"
]
}
],
"source": [
" import math #Example15_6\n",
" \n",
" \n",
" #To find the magnitude of E\n",
"k=9*10**9 #Units in N meter**2/C**2\n",
"q=3.6*10**-6 #Units in C\n",
"theta=37 #Units in degrees\n",
"r=10*math.sin(theta*math.pi/180)*10**-2 #Units in meters \n",
"e1=(k*q)/r**2 #Units in N/C\n",
"q2=5*10**-6 #Units in C\n",
"theta=37 #Units in degrees\n",
"r1=10*10**-2 #Units in meters \n",
"e2=(k*q2)/r1**2 #Units in N/C\n",
"e1y=e1 #Units in N/C\n",
"e2x=e2*math.cos(theta*math.pi/180) #Units in N/C\n",
"e2y=-e2*math.sin(theta*math.pi/180) #Units in N/C\n",
"ex=e2x #Units in N/C\n",
"ey=e1y+e2y #Units in N/C\n",
"e=math.sqrt(ex**2+ey**2) #Units in N/C\n",
"print \"The magnitude of E is=\"\n",
"print round(e,2)\n",
"print \"N/C\"\n",
" #In text book the answer isprinted wrong as E=7.26*10**6 N/C but the correct answer is E=7198876.9 N/C\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex15.7:pg-726"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The charge occured is q= 8.33e-07 C\n"
]
}
],
"source": [
" import math #Example15_7\n",
" \n",
" \n",
" #To find out how much charge occurs\n",
"e=3*10.0**6 #Units in N/C\n",
"r=0.050 #Units in meters\n",
"k=9*10.0**9 #Units in N meter**2/C**2\n",
"q=(e*r**2)/k #Units in C\n",
"print \"The charge occured is q=\",round(q,9),\" C\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex15.8:pg-727"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Lines of force come out of positive charge q suspended in cavity.\n",
"Cavity \n",
"surface must possess a negative charge since lines of force go and terminate on q.\n",
"Therefore a charge +q must exist on outer portions.\n"
]
}
],
"source": [
" import math #Example15_8\n",
" \n",
" \n",
" #To show using lines of force that a charge suspended with in cavity induces an equal and opposite charge on surface\n",
"print \"Lines of force come out of positive charge q suspended in cavity.\\nCavity \\nsurface must possess a negative charge since lines of force go and terminate on q.\\nTherefore a charge +q must exist on outer portions.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex15.9:pg-728"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The field goes by a speed of 47952.0 meters/sec\n"
]
}
],
"source": [
" import math #Example15_9\n",
" \n",
" #To find the speed just before the field strikes\n",
"e=6000 #Units in N/C\n",
"q=1.6*10**-19 #Units in C\n",
"f=e*q #Units in N\n",
"m=1.67*10**-27 #Units in Kg\n",
"a=f/m #Units in meters/sec**2\n",
"vo=0 #Units in meters/sec\n",
"x=2*10**-3 #Units in meters\n",
"v=math.sqrt(vo**2+(2*a*x)) #Units in meters/sec\n",
"print \"The field goes by a speed of \",round(v),\" meters/sec\"\n",
" #In text book answer printed wrong as v=48000 meters/sec the correct answer is v=47952 meters/sec \n"
]
}
],
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