91Ó°ÊÓ

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 $\left(100\widehat{i}\right)\text{N/C}$

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

$\begin{array}{c}\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{=}\mathbf{}{\mathbf{E}}_{\mathbf{x}}\widehat{\mathbf{i}}\mathbf{+}\mathbf{}{\mathbf{E}}_{\mathbf{y}}\widehat{\mathbf{j}}\\ \mathbf{=}\mathbf{}\frac{\mathbf{q}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Ï\mu }}_{\mathbf{o}}}\frac{\mathbf{(}\mathbf{x}\mathbf{−}{\mathbf{x}}_{\mathbf{o}}\mathbf{)}\widehat{\mathbf{i}}\mathbf{+}\mathbf{}\mathbf{(}\mathbf{y}\mathbf{−}{\mathbf{y}}_{\mathbf{o}}\mathbf{)}\widehat{\mathbf{j}}}{{\mathbf{[}{\mathbf{(}\mathbf{x}\mathbf{−}{\mathbf{x}}_{\mathbf{o}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{y}\mathbf{−}{\mathbf{y}}_{\mathbf{o}}\mathbf{)}}^{\mathbf{2}}\mathbf{}\mathbf{]}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}\end{array}$

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,

$\stackrel{→}{\left|E\right|}=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}\frac{q}{r^2}$ (ii)

The fact that the second measurement at the location$(2.0\text{cm},0)$gives$\stackrel{→}{E}=\left(100\text{N/C}\right)\widehat{i}$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:

$\stackrel{→}{E}=\frac{q}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}\frac{(x−x_o)\widehat{i}+\left(y\right)\widehat{j}}{{[{(x−x_o)}^2+{\left(y\right)}^2]}^{3/2}}$.

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

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

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

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:

$\begin{array}{c}q=4\mathrm{Ï€}{\mathrm{Ï\mu }}_o\left|\stackrel{→}{E}\right|r^2\\ =\frac{(100\text{N/C}){(0.030\text{m})}^2}{8.99×{10}^9{\text{N.m}}^{\text{2}}{\text{/C}}^{\text{2}}}\\ =1.0×{10}^{−11}\text{C}\end{array}$

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

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