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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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)
```

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"
```

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"
```

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"
```

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"
```

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
Cad=Cac*Ccd/(Cac+Ccd)
q=Cad*V
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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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)
```

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"
```

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"
```

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"
```