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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 16:Electric Potential"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex16.1:pg-731"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The magnitude of electric field is E= 2400.0 V/meters\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example16_1\n",
+ " \n",
+ " \n",
+ "#To find the magnitude of the electric field\n",
+ "v=12.0 #Units in V\n",
+ "d=5.0*10**-3 #units in Meters\n",
+ "e=v/d #Units in V/meter\n",
+ "print \"The magnitude of electric field is E=\",round(e),\" V/meters\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex16.2:pg-731"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The speed of the proton is Vab= 92858.79 meters/sec\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example16_2\n",
+ " \n",
+ " \n",
+ "#To calculate the speed of the proton\n",
+ "q=1.6*10**-19 #Units in C\n",
+ "vab=45 #Units in V\n",
+ "m=1.67*10**-27 #Units in Kg\n",
+ "va=math.sqrt((2*q*vab)/m) #Units in meters/sec\n",
+ "print \"The speed of the proton is Vab=\",round(va,2),\" meters/sec\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex16.3:pg-733"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The speed of the electron is Vab= 3975777.37 meters/sec\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example16_3\n",
+ " \n",
+ "#To find the sped of an electron\n",
+ "e=1.6*10**-19 #Units in C\n",
+ "vab=45 #Units in V\n",
+ "m=9.11*10**-31 #Units in Kg\n",
+ "va=math.sqrt((2*e*vab)/m) #Units in meters/sec\n",
+ "print \"The speed of the electron is Vab=\",round(va,2),\" meters/sec\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex16.4:pg-735"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Charged metal object have an equi potential volume so its surface is am equi potential volume\n",
+ "Because lines of force must be perpendiculat ro equipotential lines and surfaces.\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example16_4\n",
+ " \n",
+ " \n",
+ " #To sketch the equipotentials and electric field lines near a charged metal object\n",
+ "print \"Charged metal object have an equi potential volume so its surface is am equi potential volume\\nBecause lines of force must be perpendiculat ro equipotential lines and surfaces.\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex16.5:pg-735"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The work done in carrying proton is=\n",
+ "1.44e-18\n",
+ "Joules\n",
+ "\n",
+ "The work done in carrying electron is=\n",
+ "-1.44e-18\n",
+ "Joules\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example16_5\n",
+ " \n",
+ " \n",
+ "#To find the work done in carrying a proton and for an electron\n",
+ "q=1.6*10**-19 #Units in C\n",
+ "vab=9.0 #Units in V\n",
+ "work=q*vab #Units in J\n",
+ "print \"The work done in carrying proton is=\"\n",
+ "print work\n",
+ "print \"Joules\\n\"\n",
+ "q=-1.6*10**-19 #Units in C\n",
+ "work=q*vab #Units in J\n",
+ "print \"The work done in carrying electron is=\"\n",
+ "print work\n",
+ "print \"Joules\\n\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex16.6:pg-736"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The speed of proton before it strikes is Vb= 7756781.9 meters/sec\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example16_6\n",
+ " \n",
+ " #To calculate the speed just befor it strikes it\n",
+ "va=8*10**6 #Units in meters/sec\n",
+ "q=1.6*10**-19 #Units in C\n",
+ "m=1.67*10**-27 #Units in Kg\n",
+ "vab=20000 #Units in V\n",
+ "vb=math.sqrt(va**2-((2*q*vab)/m)) #Units in meters/sec\n",
+ "print \"The speed of proton before it strikes is Vb=\",round(vb,1),\" meters/sec\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex16.7:pg-737"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Since each proton has a minimum energy of 13.6 eV and a charge of 1.602*10**-19 C\n",
+ " The required potential difference is=13.6 eV\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example16_7\n",
+ " \n",
+ " \n",
+ " #To calculate the minimum value of Vab needed\n",
+ "print \"Since each proton has a minimum energy of 13.6 eV and a charge of 1.602*10**-19 C\\n The required potential difference is=13.6 eV\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex16.8:pg-737"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The speed of electron is v= 4150286.78 meters/sec\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example16_8\n",
+ " \n",
+ " #To find out the speed of the proton\n",
+ "k=9*10**9 #Units in N meter**2/C**2\n",
+ "q=5*10.0**-6 #Units in C\n",
+ "r=0.5 #Units in meters\n",
+ "v1=(k*q)/r #Units in V\n",
+ "q=1.6*10**-19 #Units in V\n",
+ "m=1.672*10**-27 #Units in Kg\n",
+ "v=math.sqrt((v1*q*2)/m) #Units in V\n",
+ "print \"The speed of electron is v=\",round(v,2),\" meters/sec\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex16.9:pg-739"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Due to 5*10**-8 C V1= 4500.0 V\n",
+ "Due to 8*10**-8 C V2= 7200.0 V\n",
+ "Due to 40*10**-8 C V3= -18000.0 V\n",
+ " Absolute potential at B is Vb= -6300.0 V\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example16_9\n",
+ " \n",
+ " \n",
+ " #To compute the absolute potential at B\n",
+ "k=9*10**9 #Units in N meter**2/C**2\n",
+ "q=5*10.0**-8 #Units in C\n",
+ "r=0.1 #Units in meters\n",
+ "v1=(k*q)/r #Units in V\n",
+ "q=8*10**-8 #Units in C\n",
+ "r=0.1 #Units in meters\n",
+ "v2=(k*q)/r #Units in V\n",
+ "q=40*10**-8 #Units in C\n",
+ "r=0.2 #Units in meters\n",
+ "v3=-(k*q)/r #Units in V\n",
+ "vb=v1+v2+v3 #Units in V \n",
+ "print \"Due to 5*10**-8 C V1=\",round(v1),\" V\\nDue to 8*10**-8 C V2=\",round(v2),\" V\\nDue to 40*10**-8 C V3=\",round(v3),\" V\\n Absolute potential at B is Vb=\",round(vb),\" V\"\n",
+ " \n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex16.10:pg-739"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The absolute potential is V= 27.2 V\n",
+ "\n",
+ "The energy that is required is W=\n",
+ "4.35e-18 J\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example16_10\n",
+ " \n",
+ " \n",
+ " #To find the absolute potential and how much energy is needed to pull the electrons from atom\n",
+ "k=9*10**9 #Units in N meter**2/C**2\n",
+ "q=1.6*10**-19 #Units in C\n",
+ "r=5.3*10**-11 #Units in meters\n",
+ "v=(k*q)/r #Units in V\n",
+ "print \"The absolute potential is V=\",round(v,1),\" V\\n\"\n",
+ "Vinfinity=0 #Units in V\n",
+ "deltaV=Vinfinity-v #Units in V\n",
+ "work=-q*deltaV #Units in J\n",
+ "print \"The energy that is required is W=\"\n",
+ "print round(work,20),\"J\" \n"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.11"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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