Question

Engineering Electromagnetics. How about solution?

Answer 1

Potential at a distance `r` from a point charge `Q` is given as

`V=Q/(4piepsilon_0r)` ......(1)

Potential for the uniform sheet of charge of surface charge density `sigma` can be found by taking integral of its electric field at a distance `z` from the distribution.

`E=sigma/(2epsilon_0)=-(delV)/(delz)`
`=>V=-int\ sigma/(2epsilon_0)dz`
`=>V=- sigma/(2epsilon_0)z+C` ........(2)
where `C` is constant of integration

On similar lines potential for uniform line charge density `lambda` can be found by taking integral of its electric field at a distance `rho` from the distribution.

`E=lambda/(2piepsilon_0rho)=-(delV)/(delrho)`
`=>V=-int\ lambda/(2piepsilon_0)1/rhodrho`
`=>V=- lambda/(2piepsilon_0)lnrho+C_1` ........(3)
where `C_1` is constant of integration

The total potential is the sum of three. Therefore we get

`V_T=Q/(4piepsilon_0r)- sigma/(2epsilon_0)z- lambda/(2piepsilon_0)lnrho+C_2` ......(4)
where we have combined constant of integration.

To calculate `C_2`, it is given that at `M(0,0,5)` `V_T=0`. Now for this point `r=sqrt(2^2+0^2+(-5)^2)=sqrt29`,
`z=5-0=5` and `rho=5-4=1`. Inserting these values in (4) we get

`0=(2xx10^-6)/(4piepsilon_0sqrt29)- (5xx10^-9)/(2epsilon_0)xx5- (8xx10^-9)/(2piepsilon_0)ln1+C_2`
Inserting value of `epsilon_0 =8.85×10^-12\ F⋅m^-1` in above we get
`C_2=-(2xx10^-6)/(4pixx8.85×10^-12sqrt29)+ (5xx5xx10^-9)/(2xx8.85×10^-12)`
`C_2=-3.34xx10^3+ 1.41xx10^3`
`C_2=-1.93xx10^3`

Inserting various values in (4) we get

`V_T=(2xx10^-6)/(4pixx8.85×10^-12r)- (5xx10^-9)/(2xx8.85×10^-12)z- (8xx10^-9)/(2pixx8.85×10^-12)lnrho-1.93xx10^3` ........(5)

For point `N(1,2,3)`
`r=sqrt((2-1)^2+(2-0)^2+(3-0)^2)=sqrt14`, `z=3-0=3` and `rho=sqrt((1-0)^2+(3-4)^2)=sqrt2`. Inserting these values in (5) we get

`V_T(N)=(2xx10^-6)/(4pixx8.85×10^-12sqrt14)- (3xx5xx10^-9)/(2xx8.85×10^-12)- (8xx10^-9)/(2pixx8.85×10^-12)lnsqrt2-1.93xx10^3`
`=>V_T(N)=(4.81- 0.85- 0.05-1.93)xx10^3`
`=>V_T(N)=1.98xx10^3\ V`