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Found in: Page 697

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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

The potential V is positive, and the sign of the charge Q is also positive which is obtained as: ${\mathrm{Q}}{=}{8}{.}{33}{×}{{10}}^{-3}{\mathrm{C}}$.

See the step by step solution

## Step 1: Given Data

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

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= ${\mathbf{k}}{\mathbf{=}}{\mathbf{9}}{\mathbf{.}}{\mathbf{00}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{9}}}{\mathbf{N}}{\mathbf{.}}\frac{{\mathbf{m}}^{\mathbf{2}}}{{\mathbf{C}}^{\mathbf{2}}}$.

## Step 5: Evaluation of the charge

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

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

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

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