Q5P - Fundamentals Of Physics

Question

A charged particle produces an electric field with a magnitude of $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{ }\mathbf{N}\mathbf{/}\mathbf{C}\mathbf{ }$ at a point that is $\mathbf{ }\mathbf{50}\mathbf{ }\mathbf{cm}$ away from the particle. What is the magnitude of the particle’s charge?

Step-by-Step Solution

Answer

The magnitude of the particle’s charge is $5.6\times {10}^{-11}\mathrm{C}$

Step 1: The given data

a)  The magnitude of the electric field, $\mathrm{E}=2.0 \mathrm{N}/\mathrm{C}$

b)  The distance of the point from the particle, $\mathrm{R}=50\hspace{0.33em}\mathrm{cm} \mathrm{or} 0.5 \mathrm{m}$

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, and q = charge of the particle

Step 3: Calculation of the charge of the particle

Using the formula of the electric field of equation (i) and the given data, we can get the charge of the particle as follows:

$$\begin{gathered} \left|\mathrm{q}\right|=4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2\mathrm{E} \\ =\frac{\left(2.00 \mathrm{N}/\mathrm{C}\right){\left(0.50 \mathrm{m}\right)}^2}{\left(8.99\times {10}^{9 }\mathrm{N}.{\displaystyle \frac{{\mathrm{m}}^2}{{\mathrm{C}}^2}}\right)} \\ =5.6\times {10}^{-11}\hspace{0.33em}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is $5.6\times {10}^{-11}\hspace{0.33em}\mathrm{C}$