Determine the potential at a point 0.50 m (i) from a`+20 muC` potential charge (ii) from a`-20 muC` point charge.

Explanation

The potential due to a point is given by V `=(KQ)/(r ) `
(a ) At a distance of 0.50 m from a a positive 20 `mu` C charge the potential is
`V= (KQ)/(r) = (9.0xx10^(9)N.m^(2)//C^(2))((20xx10^(-6)C)/(0.50m))=3.6xx10^(5)`V.
(b) For the negative charge
`V=(9.0xx10^(9)N.m^(2)//C^(2))((-20xx10^(-6)C)/(0.50m))=-36xx10^(5)`V.