What are the sign and magnitude of a point charge that produces a potential of \(-2.00{\rm{ }}V\) at a distance of \(1.00{\rm{ mm}}\)?

The potential of electric at a distance\({\rm{r}}\)from the point of charge is\({\rm{ }}V = {\rm{ }} - 200{\rm{ }}V\).

The value of\(\begin{array}{c}{\rm{ }}r = {\rm{ }}1.00{\rm{ }}mm\\ = (\frac{{1m}}{{1000mm}})\\ = 0.00100{\rm{ }}m\end{array}\)

A physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them, is referred to as an electric field.

A single charge's electric potential is as follows: to ascertain the electric potential caused by a single charge source at a particular place. The electric potential energy \({U_{Qq}}\) of a system including the test charge and the source charge that generates the field is measured after placing a test charge there. The electric potential there is then equal to the ratio given by:

\(\begin{array}{c}V = \frac{{{U_{Qq}}}}{q}\\ = \frac{{kQ}}{r}\end{array}\)

Here, the value is the Coulomb's constant, a proportionality constant. The joule/coulomb is referred to as the unit of electric potential and is referred to as the volt \((V)\).

The equation to find the distance from the point charge is:

\(V = \frac{{kq}}{r}\)

The value of \(r\) is then obtained as: \(q = \frac{{rV}}{k}\)

Then, putting the values and getting the results as:

\(\begin{array}{c}q = \frac{{(0.00100{\rm{ }}m)( - 200V)}}{{8.99 \times {{10}^9}N.{m^2}/C}}\\ = - 2.22 \times {10^{ - 13}}C\end{array}\)

Therefore, magnitude of the point charge is: \(2.22{\rm{ }}x{\rm{ }}{10^{ - 13}}{\rm{ }}C\) and is a negative charge.