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"source": [
"# Chapter 16 : Electrical Energy & Capacitance"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 16.1 Page No : 533"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {
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"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The value of E = 4000.00 v/m\n"
]
}
],
"source": [
"v_bminusv_a=-12\n",
"d=0.3*10**-2#in m\n",
"E=-(v_bminusv_a)/d\n",
"print \"The value of E = %0.2f v/m\"%E"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 16.2 Page No : 533"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"solution a\n",
"Electric potential from A to B = -40000.00 V\n",
"solution b\n",
"Change in electric potential = -0.00 joules\n",
"velocity = 2768514.16 m/s\n"
]
}
],
"source": [
"from math import sqrt\n",
"print \"solution a\"\n",
"E=8*10**4#in V/m\n",
"d=0.5#in m\n",
"delta_V=-E*d\n",
"print \"Electric potential from A to B = %0.2f V\"%delta_V\n",
"print \"solution b\"\n",
"q=1.6*10**-19#in C\n",
"delta_PE=q*delta_V\n",
"print \"Change in electric potential = %0.2f joules\"%delta_PE\n",
"m_p=1.67*10**-27#in kg\n",
"vf=sqrt((2*-delta_PE)/m_p)\n",
"print \"velocity = %0.2f m/s\"%vf"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 16.3 Page No: 534"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Solution a\n",
"Magnitude of V1 = 112375.00 v\n",
"Magnitude of V2 = -35960.00 v\n",
"solution b\n",
"Magnitude of Vp = 76415.00 v\n",
"work done = 0.31 Joule\n"
]
}
],
"source": [
"k_e=8.99*10**9 #N.m**2/c**2\n",
"q1=5*10**-6# in C\n",
"q2=-2*10**-6#in C\n",
"r1=0.4\n",
"r2=0.5\n",
"V1=(k_e*q1)/(r1)\n",
"V2=(k_e*q2)/(r2)\n",
"print \"Solution a\"\n",
"print \"Magnitude of V1 = %0.2f v\"%V1\n",
"print \"Magnitude of V2 = %0.2f v\"%V2\n",
"print \"solution b\"\n",
"vp=V1+V2\n",
"print \"Magnitude of Vp = %0.2f v\"%vp\n",
"q3=4*10**-6#in C\n",
"w=vp*q3\n",
"print \"work done = %0.2f Joule\"%w"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 16.4 Page No: 535"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Capacitance = 1.77e-12 farad\n"
]
}
],
"source": [
"e0=8.85*10**-12#in c2/N.m2\n",
"A=2*10**-4#in m2\n",
"d=1*10**-3#in m\n",
"c=(e0*A)/d\n",
"print \"Capacitance = %0.2e farad\"%c"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 16.5 Page No : 535"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"capacitance = 4.50e-05 farad\n",
"voltage between battery = 2.16e-04 c\n"
]
}
],
"source": [
"c1=3*10**-6\n",
"c2=6*10**-6\n",
"c3=12*10**-6\n",
"c4=24*10**-6\n",
"delta_v=18\n",
"c_eq=c1+c2+c3+c4\n",
"print \"capacitance = %0.2e farad\"%c_eq\n",
"q=delta_v*c3\n",
"print \"voltage between battery = %0.2e c\"%q"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 16.6 Page No : 536"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"solution a\n",
"capacitance = 1.60e-06 farad\n",
"solution b\n",
"voltage between battery = 2.88e-05 c\n"
]
}
],
"source": [
"c1=3*10**-6\n",
"c2=6*10**-6\n",
"c3=12*10**-6\n",
"c4=24*10**-6\n",
"delta_v=18\n",
"print \"solution a\"\n",
"c_eq=1/((1/c1)+(1/c2)+(1/c3)+(1/c4))\n",
"print \"capacitance = %0.2e farad\"%c_eq\n",
"q=delta_v*c_eq\n",
"print \"solution b\"\n",
"print \"voltage between battery = %0.2e c\"%q"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 16.7 Page No: 536"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"solution a\n",
"capacitance = 2.00e-06 farad\n"
]
}
],
"source": [
"c1=4*10**-6\n",
"c2=4*10**-6\n",
"print \"solution a\"\n",
"c_eq=1/((1/c1)+(1/c2))\n",
"print \"capacitance = %0.2e farad\"%c_eq"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 16.8 Page No: 537"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"solution a\n",
"Energy stored = 4671 volt\n",
"solution b\n",
"power = 240000 watt\n"
]
}
],
"source": [
"Energy=1.2*10**3#in J\n",
"c=1.1*10**-4#in f\n",
"delta_v=sqrt((2*Energy)/c)\n",
"print \"solution a\"\n",
"print \"Energy stored = %0.f volt\"%delta_v\n",
"print \"solution b\"\n",
"Energy_deliverd=600#in j\n",
"delta_t=2.5*10**-3#in s\n",
"p=(Energy_deliverd)/delta_t\n",
"print \"power = %0.f watt\"%p"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 16.9 Page No: 538"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"solution a\n",
"Capacitance = 1.96e-11 farad\n",
"solution b\n",
"Voltage = 16000.0 volt\n",
"Maximum charge = 3.14e-07 columb\n"
]
}
],
"source": [
"k=3.7\n",
"e0=8.85*10**-12#in c2/N.m2\n",
"A=6*10**-4#in m2\n",
"d=1*10**-3#in m\n",
"c=(k*e0*A)/d\n",
"print \"solution a\"\n",
"print \"Capacitance = %0.2e farad\"%c\n",
"print \"solution b\"\n",
"E_max=16*10**6#in v/m\n",
"delta_v_max=E_max*d\n",
"print \"Voltage = %0.1f volt\"%delta_v_max\n",
"Q_max=delta_v_max*c\n",
"print \"Maximum charge = %0.2e columb\"%Q_max"
]
}
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