Question-1:A Parallel Plate capacitor has following dimensions
Distance between the plates=10 cm
Area of Plate=2 m²
Charge on each plate=8.85*10^-10 C
Calculate following

1. Electric Field outside the plates
2. Electric Field Between the plates
3. Capacitance of the capacitor
4. Energy stored in the capacitor

ε₀=8.854*10^-12 C²N^-1m^-2
Solution-1:

As we know that Electric field outside the plates are zero
Electric field Inside the plates is
E=Q/ε₀A
=50NC^-1

Capacitance=ε₀A/d =8.854*10^-12 *2/.1=17.6*10^-11 F

Energy stored in
capacitor=(1/2)Q²/C=.5*8.85*10^-10*8.85*10^-10/17.6*10^-11
=22.125*10^-10 J



Question-2
Two equally charges ball are placed at a distance of 3 cm .They repel each other with a force of 4*10^-5 N.
Calculate the charge on each ball
ε₀=8.854*10^-12 C²N^-1m^-2

Solution-2
Given r=3 cm=.03m
F=4*10^-5 N
Let q be the charge
As per coulumb law
F=(1/4πε₀)(q²/r²)

or q=2*10^-9 C or -2*10^-9 C

Question-3:
Two capacitor A and B are there with capacitance 6 F and 12 F respectively
Find out the capacitance of combination when A and B are connected in parallel
Find out the capacitance of combination when A and B are connected in series

Solution-3
Given
C₁=6
C₂=12

For parallel
C=C₁+C₂=6+12=18 F

For series
(1/C)=(1/C₁)+(1/C₂)
or C=4 F

Question:-4
A point charged q is placed at the origin of the coordinate system.Let i,j,k are the unit vector along the x,y,z direction.Find
the electric Field at the point(0,3,4)

Solution-4:

Distance (r) from origin=(x²+y²+z²)^1/2
=5 m

Electric Field is given by
E = (1/4πε₀) (qr/r³)
or
E=(q/4πε₀)(3j+4k)/5³
=(q/500πε₀)(3j+4k)