Q6P - Fundamentals Of Physics

Question

What is the magnitude of a point charge that would create an electric field of $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{N}\mathbf{/}\mathbf{C}$ at points $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\hspace{0.33em}}\mathbf{m}$ away?

Step-by-Step Solution

Answer

The magnitude of the point charge is $1.11\times {10}^{-10}\mathrm{C}$

Step 1: The given data

a)  The magnitude of the electric field, $\mathrm{E}=1.00 \mathrm{N}/\mathrm{C}$

b)  The distance of the point from the particle, $\mathrm{R}=1\hspace{0.33em}\mathrm{m}$

Step 2: Understanding the concept of electric field

Using the concept of the electric field at a given point, we can get its charge value from the given formula.

 

Formula:

The magnitude of the electric field, $\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2}$                                                       (1)

where  R = The distance of field point from the charge, q = charge of the particle

Step 3: Calculation of the charge of the point

Using the given data in equation (1), the magnitude of the charge of the point is given as follows:

 $$\begin{gathered} \left|\mathrm{q}\right|=4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2\mathrm{E} \\ =\frac{\left(1.00 \mathrm{N}/\mathrm{C}\right){\left(1.00 \mathrm{m}\right)}^2}{\left(8.99\times {10}^9 \mathrm{N}.{\displaystyle \frac{{\mathrm{m}}^2}{{\mathrm{C}}^2}}\right)} \\ =1.11\times {10}^{-10} \mathrm{C} \end{gathered}$$

Hence, the value of the charge is $1.11\times {10}^{-10}\mathrm{C}$