# Chapter 4 Capacitance¶

## Example 4.1 Page no 159¶

In [5]:
#Given
m=9*10**9
r=6.4*10**6              #m

#Calculation
C=r/m

#Result
print"The capacitance of the earth is ", round(C*10**6,0),"micro F"

The capacitance of the earth is  711.0 micro F


## Example 4.2 Page no 160¶

In [21]:
#Given
m=9*10**9
c=50*10**-12
V=10**4

#Calculation
r=(m*c)*10**2
q=(c*V)

#Result
print"(i) Radius of a isolated sphere is ",r,"cm"
print"(ii) Charge of a isolated sphere is ", q*10**6,"micro C"

(i) Radius of a isolated sphere is  45.0 cm
(ii) Charge of a isolated sphere is  0.5 micro C


## Example 4.3 Page no 160¶

In [33]:
#Given
r=3*10**-3                 #m
m=9*10**9
q1=27*10**-12              #C

#Calculation
R=3*r
C=R/m
V=q1/C

#Result
print"Capacitance of the bigger drop is ", C*10**12,"pico F \npotential of the bigger drop is ",V,"Volts"

Capacitance of the bigger drop is  1.0 pico F
potential of the bigger drop is  27.0 Volts


## Example 4.4 Page no 162¶

In [37]:
#Given
m=9*10**9
ra=0.09
rb=0.1

#Calculation
C=ra*rb/(m*(rb-ra))

#Result
print"Capacitance of the capacitor is ", C*10**12,"pico F"

Capacitance of the capacitor is  100.0 pico F


## Example 4.5 Page no 162¶

In [64]:
#Given
r=2                  #cm
d=1.2

#Calculation
import math
R=(d/r)*10**2
rab=(R*2)
x=r**2+4*rab
y=math.sqrt(x)

#Result
print"ra+rb=", y,"cm \nra-rb=",r ,"cm"

ra+rb= 22.0 cm
ra-rb= 2 cm


## Example 4.6 Page no 164¶

In [69]:
#Given
d=10**-3                   #m
c=1                        #F
e=8.854*10**-12

#Calculation
A=c*d/e

#Result
print"Area is ", round(A*10**-8,1),"*10**8 m**2"

Area is  1.1 *10**8 m**2


## Example 4.7 Page no 164¶

In [74]:
#Given
A=0.02                   #m**2
r=0.5                      #m

#Calculation
import math
d=A/(4.0*math.pi*r)

#Result
print"Distance is ", round(d*10**3,2),"mm"

Distance is  3.18 mm


## Example 4.8 Page no 164¶

In [87]:
#Given
e=8.854*10**-12
K=6
A=30
d=2.0*10**-3
E=500

#Calculation
C=e*K*A/d
V=E*d*10**3
q=C*V

#Result
print"Capacitance of a parallel plate ", round(q*10**3,3),"micro C"

Capacitance of a parallel plate  0.797 micro C


## Example 4.9 Page no 165¶

In [101]:
#Given
C=300*10**-12
V=10*10**3
A=0.01
d=1*10**-3

#Calculation
q=C*V
a=q/A
E=V/d

#Result
print"(i) Charge on each plate is ", q,"C"
print"(ii) Electric flux density is ", a*10**4,"10**-4 C/m**2"
print"(iii) Potential gradient is ", E,"V/m"

(i) Charge on each plate is  3e-06 C
(ii) Electric flux density is  3.0 10**-4 C/m**2
(iii) Potential gradient is  10000000.0 V/m


## Example 4.10 Page no 165¶

In [104]:
#Given
A2=500                   #cm**2
A1=100                    #cm**2
d1=0.05                    #cm

#Calculation
d2=A2*d1/A1

#Result
print"Distance between the plates of second capacitor is ", d2,"cm"

Distance between the plates of second capacitor is  0.25 cm


## Example 4.11 page no 167¶

In [4]:
#Given
c1=0.5                  #micro F
c2=0.3                  #micro F
c3=0.2                  #micro F

#Calculation
Cp=c1+c2+c3
Cs=(1/c1)+(1/c2)+(1/c3)

#Result
print" The ratio ofmaximum capacitance to minimum capacitance is ",round (Cs,1)

 The ratio ofmaximum capacitance to minimum capacitance is  10.3


## Example 4.12 Page no 168¶

In [30]:
#Given
c1=15.0                    #micro F
c2=20.0                    #micro F
V=10**-6
v1=600                     #V

#Calculation
Cs=c1*c2/(c1+c2)
Q=Cs*V*v1
Pd=(Q/c1)*10**6
Pd1=(Q/c2)*10**6

#Result
print"(i)charge on each capacitor is",round(Q *10**3,2),"10**-3 C"
print"(ii)P.D across15 micro Fcapacitor is",round (Pd,1),"V"
print"    P.D across 20 micro F is",round (Pd1,0),"V"

(i)charge on each capacitor is 5.14 10**-3 C
(ii)p.D across15 micro Fcapacitor is 342.9 V
P.D across 20 micro F is 257.0 V


## Example 4.13 page no.168¶

In [39]:
#Given
Ca=18                          #micro F
Cb=4                          #micro F

#Calculation
import math
C=Ca*Cb
C12=math.sqrt(Ca**2-4*C)
C2=2*C12

#Result
print"The capacitance of capacitor C1 is", C12,"micro F"
print"The capacitance of capacitor C2 is",C2,"micro F"

The capacitance of capacitor C1 is 6.0 micro F
The capacitance of capacitor C2 is 12.0 micro F


## Example 4.14 Page no 168¶

In [52]:
#Given
q=750*10**-6
C1=15*10**-6
V2=20.0                            #V
C3=8*10**-6

#Calculation
V1=q/C1
V=V1+V2
q3=C3*V2
q2=q-q3
C2=q2/V2

#Result
print"The value of V1 is ", V1,"V"
print"The value of V is ",V,"V"
print"The value of capacitance is",C2*10**6,"micro F"

The value of V1 is  50.0 V
The value of V is  70.0 V
The value of capacitance is 29.5 micro F


## Example 4.15 Page no 169¶

In [60]:
#Given
C2=9.0              #micro F
C3=9.0
C4=9.0
C1=3
V=10                   #V

#Calculation
C=1/((1/C2)+(1/C3)+(1/C4))
Cab=C1+C
q=Cab*V

#Result
print"Equivalent capacitance between point A and B is ", Cab,"micro F"

Equivalent capacitance between point A and B is  6.0 micro F


## Example 4.17 Page no 169¶

In [84]:
#Given
Cab=10                         #micro F
C1=8.0                           #micro F
C2=8.0
C3=8
C4=8
C5=12
V=400

#Calculation
Cbc=((C1*C2)/(C1+C2))+C3+C4
Cac=Cab*Cbc/(Cab+Cbc)
Ccd=C1+C5
Vcd=q/Ccd
q1=C5*Vcd

#Result
print"(i) The equivalent capacitance between A and D is ", Cad,"micro f"
print"(ii) The charge on 12 micro F capacitor is ",q1*10**-3,"mC"

(i) The equivalent capacitance between A and D is  5.0 micro f
(ii) The charge on 12 micro F capacitor is  1.2 mC


## Example 4.20 Page no 171¶

In [88]:
#Given
C1=5                  #micro F
C2=6                  #micro F
V=10                  #V

#Calculation
Cp=C1+C2
q=Cp*V

#Result
print"Charge supplied by battery is ", q,"micro F"

Charge supplied by battery is  110 micro F


## Example 4.21 Page no 171¶

In [93]:
#Given
C1=2             #micro F
C2=2             #micro F
C3=2
C4=2

#Calculation
Cs=C1*C2/(C1+C2)
Cab=C3*C4/(C3+C4)

#Result
print"The capacitance of the Capacitors", Cab,"micro F"

The capacitance of the Capacitors 1 micro F


## Example 4.22 Page no 171¶

In [115]:
#Given
C1=10.0              #micro F
C2=10.0
C3=10.0
C4=10*10**-3
V=500              #V

#Calculation
Cs=1/((1/C1)+(1/C2)+(1/C3))
Cab=Cs+(C4*10**3)
Q=(C1*(500/3.0))*10**-3
Q1=C4*V

#Result
print"(a) The equivalent capacitance of the network is",round(Cab,1),"micro F"
print "(b) The charge on 12 micro F Capacitor is",Q1,"*10**-3 C"

(a) The equivalent capacitance of the network is 13.3 micro F
(b) The charge on 12 micro F Capacitor is 5.0 *10**-3 C


## Example 4.23 Page no 172¶

In [129]:
#Given
C4=6                    #micro F
C5=12
C1=8.0
C7=1

#Calculation
Cs=C4*C5/(C4+C5)
C11=(C1*Cs)/(C1+Cs)
Cs1=C1*C7/(C1+C7)
Cp=C11+Cs1
C=1/(1-(1/Cp))

#Result
print"The value of capacitance C is ", round(C,2),"micro F"

The value of capacitance C is  1.39 micro F


## Example 4.24 Page no 175¶

In [138]:
#Given
K=5
l=0.2
c=10**-9               #F
b=15.4
a=15
pd=5000                  #V

#Calculation
import math
C=(K*l*c)/(41.1*math.log10(b/a))

#Result
print"(i) The capacitance of cylindrical capacitor is ", round(C*10**9,1)*10**-9,"F"
print"(ii) The potential of the inner cylinder is equal to p.d. between two cylinders i.e potentila of inner cylinder is",pd,"V"

(i) The capacitance of cylindrical capacitor is  2.1e-09 F
(ii) The potential of the inner cylinder is equal to p.d. between two cylinders i.e potentila of inner cylinder is 5000 V


## Example 4.25 Page no 179¶

In [143]:
#Given
C=5*10**-6
V=100
C1=3*10**-6

#Calculation
q=C*V
Cp=C+C1
pd=q/Cp

#Result
print"P.D across the capacitor is ", pd,"V"

P.D across the capacitor is  62.5 V


## Example 4.26 Page no 179¶

In [156]:
#Given
V=250              #V
C1=6                #micro F
C2=4
Cp=10*10**-6

#Calculation
pd=V*C1/(C1+C2)
q=pd*C2*10**-6
q1=2*q
pd1=q1/Cp
q2=C2*pd1
q3=C1*pd1

#Result
print"New potentila difference is ", pd1,"V"
print"Charge on 4 micro F capacitor is ",q2,"micro C"
print"Charge on 6 micro F capacitor is ",q3,"micro C"

New potentila difference is  120.0 V
Charge on 4 micro F capacitor is  480.0 micro C
Charge on 6 micro F capacitor is  720.0 micro C


## Example 4.28 Page no 180¶

In [10]:
#Given
C1=16*10**-6                    # F
C2=4                      #micro F
V1=100                   #V
Cp=20*10**-6               #f

#Calculation
q=C1*V1
U1=0.5*C1*V1**2
V=q/Cp
U2=0.5*Cp*V**2

#Result
print"(i) Potential difference across the capacitor is ",  V,"Volts"
print"(ii) The electrostatic energies before and after the capacitors are connected ",U2,"J"

(i) Potential difference across the capacitor is  80.0 Volts
(ii) The electrostatic energies before and after the capacitors are connected  0.064 J


## Example 4.29 Page no 180¶

In [19]:
#Given
m=9*10**9
V=3.0*10**6
r=2

#Calculation
q=(V*r)/m
E=0.5*q*V

#Result
print"The heat generated is ", E,"J"

The heat generated is  1000.0 J


## Example 4.30 Page no 180¶

In [26]:
#Given
V=12                  #V
C=1.35*10**-10          #C

#Calculation
q=C

#Result
print"Extra Charge supplied by battery is ", q,"C"

Extra Charge supplied by battery is  1.35e-10 C


## Example 4.31 Page no 181¶

In [31]:
#Given
C=100*10**-6             #F
V=500                    #V

#Calculation
q=V/2.0
E=0.5*(0.5*C*V**2)

#Result
print"Charge in the new stored energy is ", E,"J"

Charge in the new stored energy is  6.25 J


## Example 4.32 Page no 187¶

In [37]:
#Given
A=2*10**-3                    #m**2
d=0.01                        #m
t=6*10**-3                    #m
K=3
a=8.854*10**-12

#Calculation
C=a*A/(d-t*(1-(1/3.0)))

#Result
print"The capacitance of the capacitor is ", round(C*10**12,2)*10**-12,"F"

The capacitance of the capacitor is  2.95e-12 F


## Example 4.33 Page no 188¶

In [52]:
#Given
e=8.854*10**-12
A=2
t1=0.5*10**-3
t2=1.5*10**-3
t3=0.3*10**-3
K1=2.0
K2=4.0
K3=6.0

#Calculation
C=(e*A)/((t1/K1)+(t2/K2)+(t3/K3))

#Result
print"The capacitance of the capacitor is ", round(C*10**6,3)*10**-6,"F"

The capacitance of the capacitor is  2.6e-08 F


## Example 4.34 Page no 188¶

In [56]:
#Given
a=3                  #mm
b=4.0                  #mm
K1=5

#Calaculation
K2=1/((a**2/b)-a/b)*K1

#Result
print"The relative permittivity of the additional dielectric is ", round(K2,2)

The relative permittivity of the additional dielectric is  3.33


## Example 4.35 Page no 188¶

In [62]:
#Given
d=5
t=2
K=3.0

#Calculation
D=d+(t-t/K)

#Result
print"New seperaion between the plates are ", round(D,2),"mm"

New seperaion between the plates are  6.33 mm


## Example 4.36 Page no 189¶

In [78]:
#Given
d=4
t=2
K=4.0
C1=50*10**-12                    #f
V0=200                           #V

#Calculation
C=(d-t+(t/K))/d
q=C1*V0
V=V0*C
U=0.5*q*V
E=0.5*q*(V0-V)

#Result
print"(i) Final charge on ach plate is ", q,"C"
print"(ii) P.D batween the plates is ", V,"volts"
print"(iii)Final energy in the capacitor is ", U,"J"
print"(iv) Energy loss is ", E,"J"

(i) Final charge on ach plate is  1e-08 C
(ii) P.D batween the plates is  125.0 volts
(iii)Final energy in the capacitor is  6.25e-07 J
(iv) Energy loss is  3.75e-07 J


## Example 4.39 Page no 193¶

In [82]:
#Given
V=25*10**5
E=5.0*10**7

#Calculation
r=V/E

#Result
print"Minimum radius of the spherical shell is ", r*100,"cm"

Minimum radius of the spherical shell is  5.0 cm