Q29PE - College Physics (Urone)

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?

Step-by-Step Solution

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 1: Given Data

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

Distance from the charge is 15.0 m.

Step 2: 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.

Step 3: 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.

Step 4: 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= $\mathbf{k}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{N}\mathbf{.}\frac{{\mathbf{m}}^{\mathbf{2}}}{{\mathbf{C}}^{\mathbf{2}}}$.

Step 5: 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: $\mathbf{Q}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{33}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{C}$.