In the electric field of a point charge Q electric potential at a point will be

`V(vecr) = ((kQq)/r)/q = (kQ)/r`

Potential is a scalar quantity.Electric potential due to a positive charge is taken to be positive and that due to a negative charge is taken to be negative. The potential at a point due to more than one charge can be found simply by adding the potentials due each charge separately.

Consider a uniformly charged wire of infinite length.

Charge per unit length on wire: `lamda` (here assumed positive).

Electric field at radius r : `E=(2klamda)/r`

Electric potential at radius r :

`V=-2klamdaint_(r_0)^r1/r dr = -2klamda[lnr-lnr_0]`

`=>V=2klamdaln(r_0)/r`

Here we have used a finite, nonzero reference radius `r_0 ne 0, oo.`

The illustration from the textbook uses `R_(r e f)` for the reference radius, R for the integration variable, and `R_p` for the radial position of the field point.

If the charge on the shell = q

`(i)` For `r > R`

`E=(kq)/r^2`

`=> V=-int_oo^rEdr=-int_oo^r(kq)/r^2 dr= (kq)/r`

`(ii)` For r = R, `V=(kq)/R`

`(iii)` For r < R

`E=0` (r < R)

`=> V=-int_oo^rEdr=-[int_oo^REdr+int_R^rEdr]`

`=-int_oo^r(kq)/r^2dr-int_R^r0dr=(kq)/r`

If the total charge = Q

`(i)` for r > R,

Volume charge density `rho=Q/(4/3piR^3)`

`E(r > R) = (kQ)/r^2`

`V=-int_oo^r(kQ)/r^2dr=(kQ)/r`

`(ii)` for `r=R`, `V=(kQ)/R`

`(iii)` `E( r < R) = (rho r)/(3epsilon_0)=(Qr)/(4/3piR^3 3epsilon_0`

`E(r < R) = (Qr)/(4piepsilon_0R^3)=(kQr)/R^3`

`V=-[int_oo^REdr+int_R^rEdr]=int_oo^R(kQ)/r^2 dr-int_R^r(kQr)/R^3dr`

`V=(kQ)/(2R)[3-r^2/R^2]`