In [5]:

```
#Given
q=300*10**-6 #c
V=6
#Calculation
W=q*V
#Result
print"Work done is ", W*10**3,"*10**-3 J"
```

In [9]:

```
#given
Va=-10 #V
W=300 #J
q=3.0 #C
#Calculation
V=(W/q)+Va
#Result
print"The value of V is ", V,"Volts"
```

In [17]:

```
#Given
m=9*10**9
q=16*10**-10 #C
r=0.1
r1=0.06
q1=12*10**-10
#Calculation
Vb=m*q/r
Vb1=m*q/r1
V=Vb1-Vb
W=q1*V
#Result
print"Workdone is ", W*10**8,"*10**-8 J"
```

In [22]:

```
#Given
r=3.4*10**-14 #m
n=47
q=1.6*10**-19 #C
m=9*10**9
#Calculation
V=m*n*q/r
#Result
print"Electric potential at the surface of silver nucleus is ", round(V*10**-6,2),"*10**6 V"
```

In [26]:

```
#Given
m=9*10**9
q=4*10**-6
#Calculation
V=2*q*m
#Result
print"Electric potential is ", V*10**-3,"*10**3 V"
```

In [30]:

```
#Given
m=9*10**9
q=250*10**-6
r=0.1
#Calculation
V=m*q/r
#Result
print"Electric potential at the centre is ", V*10**-7,"*10**7 V"
```

In [38]:

```
#Given
m=3*10**-16
g=9.8
d=5*10**-3
q=16.0*10**-18
#Calculation
V=(m*g*d/q)*10
#Result
print"Voltage needed to balance an oil drop is ",round(V,2),"V"
```

In [44]:

```
#Given
q=1.6*10**-19 #C
V=3000 #V
r=5*10**-2 #m
g=9.8
#Calculation
E=V/r
m=q*E/g
#Result
print"The mass of the particle is ", round(m*10**16,1),"*10**-16 Kg"
```

In [48]:

```
#Given
m=9*10**-9
q1=3*10**-9
q2=3*10**-9
q3=10**9
r=0.2
#Calculation
W=m*((q1*q3/r)+(q2*q3/r))
#Result
print"Workdone is ", W,"J"
```

In [53]:

```
#Given
m=9*10**9
q=1.6*10**-19
r=10**-10
#Calculation
U=m*q**2/r
K=U/2.0
#Result
print"Kinetic energy is ",K,"J"
```

In [57]:

```
#Given
m=9*10**-31
V=10**6
q=1.6*10**-19
a=9*10**9
#Calculation
K=m*V**2
r=a*q**2/K
#Result
print"Distance of the closest approach is ", r,"m"
```

In [75]:

```
#Given
r=0.53*10**-10 #m
q1=1.6*10**-19 #C
q2=-1.6*10**-19 #C
a=9*10**9
r1=1.06*10**-10
#Calculation
U=a*q1*q2/r
Ue=U/q1
K=-Ue/2.0
E=Ue+K
U1=(a*q1*q2/r1)/q1
#Result
print"(i) Potential energy of the system is ", round(Ue,1),"eV"
print"(ii) Minimum amount of work required to free the elctrons ia ",round(E,1),"ev"
print"(iii) Potential energyof the system is ",round(E,1) ,"ev and work requiredto free the electrons is ",round(-E,1),"eV"
```

In [93]:

```
#Given
a=9*10**9
q1=7*10**-6 #C
q2=-2*10**-6
r=0.18
r1=0.09
A=9*10**5
#Calculation
U=a*q1*q2/r
W=0-U
U1=(q1*A/r1)+(q2*A/r1)+U
#Result
print"(a) Electrostatic potential energy is ", round(U,1),"J"
print"(b) Work required to seperate two charges is ",round(W,1),"J"
print"(c) Electrostatic energy is ", U1,"J"
```

In [103]:

```
#Given
p=6*10**-6
E=10**6
a=1
#Calculation,
U1=-p*E*a
U2=(p*E*(math.cos(60)*180/3.14))*10**-2
U3=U2-U1
#Result
print"Heat released by substance is ", round(U3,0),"J"
```

In [108]:

```
#Given
q=10**-7
e=8.854*10**-12
#Calculation
a=q/e
#Result
print"Electric flux through the surface of the cube is ", round(a*10**-4,2),"Nm**2C-1"
```

In [115]:

```
#Given
q=8.85*10**-6
e=8.85*10**-12
#Calculation
a=q/e
b=a/6.0
#Result
print"Electric flux through each face is ", round(b*10**-5,2),"Nm**2C-1"
```

In [118]:

```
#Given
E0=2*10**3 #N/C
S=0.2
#Calculation
a=(3/5.0)*E0*S
#Result
print"Electric flux of the field is ", a,"Nm**2C-1"
```

In [128]:

```
#Given
r=0.2
m=9*10**9
b=50
import math
E=250*r
a=E*4*math.pi*r**2
q=b*r**2/m
#Result
print"Charge contained in a sphere is ", round(q*10**10,2)*10**-10,"C"
```

In [139]:

```
#Given
a=0.1 #m
A=800
e=8.854*10**-12
#Calculation
b=A*a**2.5*(math.sqrt(2)-1)
q=e*b
#Result
print"(a) The flux through the cube is ", round(b,2),"Nm**2C-1"
print"The charge within the cube is ",round(q*10**12,2)*10**-12,"C"
```

In [156]:

```
#Given
E=200
a=0.05
e=8.854*10**-12
d=3.14
#Calculation
import math
b=E*math.pi*a**2
c=2*b
q=e*d
#Result
print"(a) Net outward flux through each flat face is ", round(b,2),"Nm**2C-1"
print"(b) Flux through the side of cylinder is zero "
print"(c) Net outward flux through the cylinder is ", round(c,2),"Nm**2C-1"
print"(d) The net charge in the cylinder is ",round(q*10**11,2)*10**-11,"C"
```

In [7]:

```
#Given
q=5.8*10**-6 #C
r=8*10**-2 #m
e=8.854*10**-12
l=3.0
#Calculation
import math
E=q/(2*math.pi*e*r*l)
#Result
print"Electric field is ", round(E*10**-5,1),"*10**5 N/C"
```

In [10]:

```
#Given
E=9*10**4 #N/C
r=2*10**-2 #m
m=9*10**9
#Calculation
a=r*E/(2.0*m)
print"Linear charge density is ", a,"Cm-1"
```

In [17]:

```
#Given
q=10*10**-6 #C
r=0.1 #m
a=8.85*10**-12
#Calculation
import math
E=q/(4.0*math.pi*a*r**2)
#Result
print"(i) Electric field intensity at a point 10 cm from the centre", round(E*10**-6,0),"*10**6 N/C"
print"(ii) Since the point is lying inside the shell, electric intensity at this point is zero"
```

In [31]:

```
#Given
Z=79
e=1.6*10**-19
e0=8.854*10**-12
R=6.2*10**-15
#Calculation
import math
q=Z*e
E=q/(4.0*math.pi*e0*R**2)
b=E/4.0
#Result
print"(i) The magnitude of the electric field at the surface of nucleus is ", round(E*10**-21,0)*10**21,"N/C"
print"(ii) The magnitude of the electric field at a distance 2R from the centre of the nucleus is ",round(b*10**-21,2),"*10**21 N/C"
```

In [43]:

```
#Given
e=8.854*10**-12
A=0.5
F=1.8*10**-12 #N
E=1.6*10**-19
#Calculation
q=(2*e*A**2*F)/E
#Result
print"Total charge on the sheet is ", round(q*10**6,0),"micro C"
```

In [54]:

```
#Given
a=5*10**-6
e=8.854*10**-12
r=0.1
#Calculation
import math
b=-(a*math.pi*r**2*(math.cos(60)*180/3.14))/(2*e)
#Result
print"Electric flux through a circular area is ", round(b*10**-5,2),"*10**3 Nm**2C-1"
```