Answer 1

i) Let P be the point at which electric potential is required.

Potential at P due to − q charge,

`V_1=-q/(4piepsilon_0r_1)`

Potential at P due to + q charge,

`V_2=q/(4piepsilon_0r_2)`

Potential at P due to the dipole,

`V=V_1+V_2`

`V=q/(4piepsilon_0)[1/r_2-1/r_1]`   .....(i)

Now, by geometry

`r_1^2=r^2+a^2+2arcostheta`

`r_2^2=r^2+a^2+2arcos(180^@-theta)`
`r_2^2=r^2+a^2-2arcostheta`

`r_1^2=r^2(1+a^2/r^2+(2a)/rcostheta)`

If r >> a, `a/r` is small. Therefore,`a^2/r^2` can be neglected.

`r_1^2=r^2(1+(2a)/rcostheta)`

`1/r_1=1/r(1+(2a)/rcostheta)^(1/2)`

similarly, `1/r_2=1/r(1-(2a)/rcostheta)^(1/2)`

Putting these values in (i), we obtain

`V=q/(4piepsilon_0)[1/r(1-(2a)/rcostheta)^(1/2)-1/r(1+(2a)/rcostheta)^(1/2)]`

Using Binomial theorem and retaining terms up to the first order in`a/r`, we obtain

`V=q/(4piepsilon_0r)[1+a/rcostheta-(1-a/rcostheta)]`

`V=q/(4piepsilon_0r)[1+a/rcostheta-1+a/rcostheta]`

`V=q/(4piepsilon_0r)[(2a)/rcostheta]`

`V=(qxx2acostheta)/(4piepsilon_0r^2)`

`V=(Pcostheta)/(4piepsilon_0r^2)`

For axial line put θ=0°

`V=(Pcos(0))/(4piepsilon_0r^2)`

`V=P/(4piepsilon_0r^2)`