91影视

Given charge $=q=10.0nC=10.0脳{10}^{-9}C$

Given electric field$=\stackrel{鈫�}{E}=鈭�225,000\hat{i}N/C$

The position of the point charge is $(x,y)=(1.0cm,2.0cm)$

Hence, the diagram in space of the charged particle can be made as:

The given electric field has a negative charge, which indicates that the point is located on the left-hand side in the x-direction. The given electric field has only $\hat{i}$component, which means it has only an x-component. Also, it means that the y-component is zero, and hence its y-coordinates won't change.

The magnitude of the electric field at a point is given by the equation:

$E=\frac{1}{4蟺{系}_0}\frac{q}{r^2}$

By rearranging the terms, we get the equation of r as:

$r=\sqrt{\frac{1}{4蟺{系}_0}\frac{q}{E}}$

By substituting the value in the above equation, we get,

localid="1648546847383" $$\begin{gathered} r=\sqrt{\frac{9.0脳{10}^9脳10脳{10}^{鈭�9}}{225000}} \\ r=0.02m=2.0cm \end{gathered}$$

Hence, the x-coordinate can be calculated as:

$$\begin{gathered} x=1.0-r \\ x=1.0-2.0 \\ x=-1.0cm \end{gathered}$$

Therefore, the position is$(x,y)=(-1.0cm,2.0cm)$

Thus, the position of the given electric field is$(x,y)=(-1.0cm,2.0cm)$.

Given charge $=q=10.0nC=10.0脳{10}^{-9}C$

Given electric field$=\stackrel{鈫�}{E}=(161,000\hat{i}+80,500\hat{j})N/C$

The position of the point charge is$(x,y)=(1.0cm,2.0cm)$

Hence, the diagram in space of the charged particle can be made as:

From the given form of the electric field, it can be said that it has two components i.e. an x-component and a y-component. Hence, the net electric field can be calculated as:

$$\begin{gathered} E=\sqrt{{(x-component)}^2+{(y-component)}^2} \\ E=\sqrt{{161000}^2+{80500}^2} \\ E=180003.4N/C \end{gathered}$$

The magnitude of the electric field at a point is given by the equation:

$E=\frac{1}{4蟺{系}_0}\frac{q}{r^2}$

By rearranging the terms, we get the equation of r as:

$r=\sqrt{\frac{1}{4蟺{系}_0}\frac{q}{E}}$

By substituting the value in the above equation, we get,

$$\begin{gathered} r=\sqrt{\frac{9.0脳{10}^9脳10脳{10}^{鈭�9}}{180003.4}} \\ r=0.022m=2.2cm \end{gathered}$$

And, the angle between the position of the electric field and the point charge can be calculated as:

$$\begin{gathered} 胃={\tan }^{-1}\left(80500/161000\right) \\ 胃=26.56掳 \end{gathered}$$

Hence, the distance from x can be calculated as:

$$\begin{gathered} x=r\cos \left(胃\right) \\ x=2.2\cos \left(26.56\right) \\ x=2.0cm \end{gathered}$$

And, the distance from y can be calculated as:

$$\begin{gathered} y=r\sin \left(胃\right) \\ y=2.2\sin \left(26.56\right) \\ y=1.0cm \end{gathered}$$

Therefore, the coordinates can be calculated as:

$$\begin{gathered} (x,y)=(1.0+2.0cm,2.0+1.0cm) \\ (x,y)=(3.0cm,3.0cm) \end{gathered}$$

Thus, the position of the given electric field is$(x,y)=(3.0cm,3.0cm)$.

Given charge $=q=10.0nC=10.0脳{10}^{-9}C$

Given electric field $=\stackrel{鈫�}{E}=(21,600\hat{i}鈭�28,800\hat{j})N/C$

The position of the point charge is $(x,y)=(1.0cm,2.0cm)$

Hence, the diagram in space of the charged particle can be made as:

From the given form of the electric field, it can be said that it has two components i.e. an x-component and a y-component. Also, the y-component is negative, therefore, the position of the electric field is on the downward side of the given charge. The net electric field can be calculated as:

$$\begin{gathered} E=\sqrt{{(x-component)}^2+{(y-component)}^2} \\ E=\sqrt{{21600}^2+{28800}^2} \\ E=36000N/C \end{gathered}$$

The magnitude of the electric field at a point is given by the equation:

$E=\frac{1}{4蟺{系}_0}\frac{q}{r^2}$

By rearranging the terms, we get the equation of r as:

$r=\sqrt{\frac{1}{4蟺{系}_0}\frac{q}{E}}$

By substituting the value in the above equation, we get,

$$\begin{gathered} r=\sqrt{\frac{9.0脳{10}^9脳10脳{10}^{鈭�9}}{36000}} \\ r=0.05m=5.0cm \end{gathered}$$

And, the angle between the position of the electric field and the point charge can be calculated as:

$$\begin{gathered} 胃={\tan }^{-1}\left(28800/21600\right) \\ 胃=53.13掳 \end{gathered}$$

Hence, the distance from x can be calculated as:

$$\begin{gathered} x=r\cos \left(胃\right) \\ x=5.0\cos \left(53.13\right) \\ x=3.0cm \end{gathered}$$

And, the distance from y can be calculated as:

$$\begin{gathered} y=r\sin \left(胃\right) \\ y=5.0\sin \left(53.13\right) \\ y=4.0cm \end{gathered}$$

Therefore, the coordinates can be calculated as:

$$\begin{gathered} (x,y)=(1.0+3.0cm,2.0-4.0cm) \\ (x,y)=(4.0cm,-2.0cm) \end{gathered}$$

Thus, the position of the given electric field is $(x,y)=(4.0cm,-2.0cm)$.