Q73P - Fundamentals Of Physics

Question

The electric field in ax-y plane produced by a positively charged particle is $\mathbf{7}\mathbf{.2}\mathbf{(}\mathbf{4}\mathbf{.0}\widehat{\mathbf{i}}\mathbf{+}\mathbf{ }\mathbf{3}\mathbf{.0}\widehat{\mathbf{j}}\mathbf{)}\mathbf{ }\mathbf{\text{N/C}}$ at the point (3.0, 3.0) cm and  $\mathbf{100}\widehat{\mathbf{i}\mathbf{ }}\mathbf{\text{N/C}}$at the point (2.0, 0) cm. What are the (a) x and (b) y coordinates of the particle? (c) What is the charge of the particle?

Step-by-Step Solution

Answer

a) The x-coordinate of the particle is.$-1.0\text{cm}$

b) The y-coordinate of the particle is.$0\text{cm}$

c) The charge of the particle is. $1.0\times {10}^{-11}\text{C}$

 

Step 1: The given data

a) A positively charged particle at the point$(3.0,3.0)\text{cm}$ produces an electric field of  $7.2(4.0\widehat{i}+3.0\widehat{j}) \text{N/C}$

b) A positively charged particle at the pointproduces an electric field of  $\left(100\widehat{i}\right) \text{N/C}$

Step 2: Understanding the concept of the electric field

We have a positive charge in the x-y plane. From the electric fields it produces at two different locations, we can determine the position and the magnitude of the charge.Let the charge be placed at the point, (x, y). In Cartesian coordinates, the electric field at a point (x, y) can be written as

 $\begin{array}{c}\overrightarrow{\mathbf{E}}\mathbf{=}\mathbf{ }{\mathbf{E}}_{\mathbf{x}}\widehat{\mathbf{i}}\mathbf{+}\mathbf{ }{\mathbf{E}}_{\mathbf{y}}\widehat{\mathbf{j}}\\ \mathbf{=}\mathbf{ }\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{ϵ}}_{\mathbf{o}}}\frac{\mathbf{(}\mathbf{x}\mathbf{-}{\mathbf{x}}_{\mathbf{o}}\mathbf{)}\widehat{\mathbf{i}}\mathbf{+}\mathbf{ }\mathbf{(}\mathbf{y}\mathbf{-}{\mathbf{y}}_{\mathbf{o}}\mathbf{)}\widehat{\mathbf{j}}}{{\mathbf{[}{\mathbf{(}\mathbf{x}\mathbf{-}{\mathbf{x}}_{\mathbf{o}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{y}\mathbf{-}{\mathbf{y}}_{\mathbf{o}}\mathbf{)}}^{\mathbf{2}}\mathbf{ }\mathbf{]}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}\end{array}$

Formulae:

The ratio of the field components is given by: 

$\frac{E_y}{E_x}= \frac{y-y_o}{x-x_o}$                                                                                   (i)

The magnitude of the electric field of a particle,

$\overrightarrow{\left|E\right|}=\frac{1}{4\pi {ϵ}_o}\frac{q}{r^2}$                                                                                                            (ii)

 

Step 3: a) Calculation of the x-coordinate of the particle

The fact that the second measurement at the location $(2.0\text{ cm},0)$gives $\overrightarrow{E}=\left(100\text{ N/C}\right)\widehat{i}$ indicates that, $y_o=0$that is, the charge must be somewhere on the x axis. Thus, the above expression can be simplified to the given value:

$\overrightarrow{E}= \frac{q}{4\pi {ϵ}_o}\frac{(x-x_o)\widehat{i}+ \left(y\right)\widehat{j}}{{[{(x-x_o)}^2+{\left(y\right)}^2 ]}^{3/2}}$. 

On the other hand, the field at $(3.0\text{ cm},3.0\text{ cm})$is,$\overrightarrow{E}=\left(7.2\text{ N/C}\right)(4.0\widehat{ i}+3.0\widehat{j})$ which gives.Thus, thus, using equation (i), we have the ratio as:

Hence, the value of the x-coordinate is.$-1.0\text{cm}$

Step 4: b) Calculation of the y-coordinate of the particle

As shown in the above calculations of part (a), the y-coordinate of the particle is$0\text{cm}$$0\text{cm}$

 

Step 5: c) Calculation of the charge of the particle .

We note that the field magnitude measured at $(2.0\text{cm},0)$(which is  $r=0.030\text{m}$from the charge). Thus, the value of the charge of the particle using equation (ii) is given as:

$\begin{array}{c}q= 4\pi {ϵ}_o\left|\overrightarrow{E}\right|r^2\\ = \frac{(100 \text{N/C}){(0.030 \text{m})}^2}{8.99\times {10}^9{\text{N.m}}^{\text{2}}{\text{/C}}^{\text{2}}}\\ =1.0\times {10}^{-11}\text{C}\end{array}$

Therefore, the charge of the particle is.$1.0\times {10}^{-11}\text{C}$

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