Question

Two charges are placed in a line with a distance of 10cm. At the left end
of the line, a 12nC is placed and to the right end of the line, a –12nC is
placed. Determine the potential at the point 4cm to the left of 12nC, at
the point 4cm to the -12nC and a point 13cm from both charges.

Answer 1

$V_1=\frac{kQ}{r}$

$V_1=\frac{9\times10^9\times12\times10^{-9}}{14\times10^{-2}}=771V$

$V_2=-\frac{9\times10^9\times12\times10^{-9}}{4\times10^{-2}}=-2700V$

$$V=V_1+V_2=771-2700=-1929V$$

$V'=\frac{kQ}{r}$

$V_1=\frac{9\times10^9\times12\times10^{-9}}{13\times10^{-2}}=830V$