Chapter 11:Capacitance

Problem no:11.1, Page no:123

In [1]:
from __future__ import division

#Cal of charge on capacitors plates

#initialization

C=200                          #Capacitance in pF

V=100                          #Voltage in V

#Calculation

C=200E-12                      #Convert Capacitance in F

Q=C*V                          #Capacitance formula

print "Q=",Q,"C"

Q= 2e-08 C

Problem no:11.2, Page no:123

In [3]:
from __future__ import division

#Cal of Capacitance

#Initialization

Q=5E-4                            #Charge in C

V=300                             #Voltage in V

#Calculation

C=Q/V                             #Capacitance Formula           

C=C*(10**6)                       #Convert Capacitance in muF
 
print "C=",round(C,2),"muF"

The Capacitance is 1.67 muF

Problem no:11.3, Page no:124

In [4]:
from __future__ import division

#Cal of Capacitance

#Initialization

Eo=8.85E-12                     #Permittivity in F/m

A=5                             #Area in cm^2

d=0.1                           #distance in mm

#Calculation

K=1                             #dielectric constant in air

A=A/100                         #Convert Area in m^2

d=d/1000                        #Convert diameter in m

C=(K*Eo*(A**2))/(d)

C=int(C*(10**12))                    #Convert Capacitance in pF

print "(a)C=",C,"pF"

K=6                             #Permittivity for mica given

C=K*C

print "(b)C=",C,"pF"

(a)C= 221 pF
(b)C= 1326 pF

Problem no:11.4, Page no:124

In [5]:
from __future__ import division

#Cal of dielectric constant of Benzene

#Initialization

C1=2                        #Capacitance in air in muF

C2=4.6                      #Capacitance in Benzene

K1=1                        #Dielectric constant for air

#Calculation

K2=K1*(C2/C1)

print "K2=",K2

K2= 2.3

Problem no:11.7, Page no:125

In [14]:
from __future__ import division

import math

#Cal of voltage and charge

#Initialization

C=100                     #Capacitance in muF

W=50                      #Energy in J

#Calculation
 
C=C*(10**-6)              #Convert Capacitance in F 
 
V=math.sqrt((2*W)/C)

print "(a)V=",int(V),"V"

Q=C*V
print "(b)Q=",Q,"C"

(a)V= 1000 V
(b)Q= 0.1 C

Problem no:11.8, Page no:126

In [6]:
from __future__ import division

#Cal of Equivalent Capacitance

#Initialization

#Capacitance in muF

C1=1

C2=2

C3=3

#Calculation

C=1/((1/C1)+(1/C2)+(1/C3))      #Formula for capacitors in series

print "C=",round(C,3),"muF"

C= 0.545 muF

Problem no:11.9, Page no:126

In [7]:
from __future__ import division

#Cal of Equivalent Capacitance

#Initialization

#Capacitance in muF

C1=2

C2=3

#Calculation

C=(C1*C2)/(C1+C2)              #Formula for Capacitors in series

print "C=",C,"muF"

C= 1.2 muF

Problem no:11.12, Page no:127

In [9]:
from __future__ import division


#Initialization

#Cal of Equivalent capacitance

#Capacitance in muF

C1=5

C2=10

V=1000                   #Voltage in Volts

C=C1+C2

print "(a) C=",C,"pF"

#Cal of Charge

Q1=(C1*10**-12)*V

Q2=(C2*10**-12)*V

print "(b) Q1=",Q1,"C"

print "    Q2=",Q2,"C"

(a) C= 15 pF
(b) Q1= 5e-09 C
    Q2= 1e-08 C

Problem no:11.13, Page no:130

In [52]:
from __future__ import division


#Initialization

#Cal of Final charge

C=20                            #Capacitance in muF

V=45                            #Voltage in volts

R=2000                          #Resistance in Ohm

Qo=(C*10**-6)*V

print "(a) Qo=", "%.0e" %Qo,"C"

#Cal of Time

T=R*(C*10**-6)

print "(b) T=",T,"s"

(a) Qo= 9e-04 C
(b) T= 0.04 s

Problem no:11.14, Page no:130

In [54]:
from __future__ import division

import math

#Initialization

t=0.01

Qo=9E-4

#Cal of charge when t=0.01 s

t_T=round(math.exp(-t/0.04),2)

Q=Qo*(1-t_T)

print "Q=","%.0e" %Q,"C"


#Cal of Charge when t=0.1 s

t=0.1

t_T=round(math.exp(-t/0.04),2)

Q=Qo*(1-t_T)

print "Q=","%.1e" %Q,"C"

Q= 2e-04 C
Q= 8.3e-04 C