Question

If the potential due to a point charge is $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}}\mathbf{}\mathbf{V}$at a distance of 15.0 m, what are the sign and magnitude of the charge?

Short answer

The potential V is positive, and the sign of the charge Q is also positive which is obtained as: $\mathrm{Q}=8.33\times {10}^{-3}\mathrm{C}$.

Step-by-step solution

Given Data

Potential due to the charge is $5.00\times {10}^2\mathrm{V}$.

Distance from the charge is 15.0 m.

Define Electric Field

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

The given values

Finding the magnitude and sign of the charge Q, if the potential is due to a point charge is:$\mathrm{V}=5.00\mathrm{x}10\mathrm{V}$,

It is at a distance of r which is: r = 15.0 m.

Obtaining the value of :

In the equation of the potential V is:$\mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{r}}$

Determine the charge Q, as:$\mathrm{Q}=\frac{\mathrm{rV}}{\mathrm{k}}$

The value of k= $\mathbf{k}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{N}\mathbf{.}\frac{{\mathbf{m}}^{\mathbf{2}}}{{\mathbf{C}}^{\mathbf{2}}}$.

Evaluation of the charge

Substituting the quantities in the equation for the charge Q as:

$\begin{array}{rcl}\mathbf{Q}& \mathbf{=}& \frac{\mathbf{rV}}{\mathbf{k}}\\ & \mathbf{=}& \frac{\mathbf{(}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}}\mathbf{V}\mathbf{)}}{\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathbf{N}\mathbf{.}\frac{{\mathbf{m}}^{\mathbf{2}}}{{\mathbf{C}}^{\mathbf{2}}}}\\ & \mathbf{=}& \mathbf{8}\mathbf{.}\mathbf{33}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{C}\\ & & \end{array}$

Therefore, the potential V is positive and the sign of the charge Q is positive which gives us:$\mathbf{Q}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{33}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{C}$.