Question

A $10.0nC$charge is located at position $(x,y)=(1.0cm,2.0cm)$. At what $(x,y)$position(s) is the electric field

1. localid="1648545048454" $-225,000\hat{i}N/C$
2. localid="1648545057140" $(161,000\hat{i}+80,500\hat{j})N/C$
3. $(21,600\hat{i}-28,800\hat{j})N/C$

Short answer

1. $(x,y)=(-1.0cm,2.0cm)$
2. $(x,y)=(3.0cm,3.0cm)$
3. $(x,y)=(4.0cm,-2.0cm)$
   

Step-by-step solution

Part (a) Step 1: Given Information

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

Given electric field$=\overrightarrow{E}=-225,000\hat{i}N/C$

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:

Part (a) Step 2: Calculation

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:

$E=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}$

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

$r=\sqrt{\frac{1}{4\pi {ϵ}_0}\frac{q}{E}}$

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

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

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$(x,y)=(-1.0cm,2.0cm)$

Part (a) Step 3: Final answer

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

Part (b) Step 1: Given Information

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

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

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:

Part (b) Step 2: Calculation

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:

$E=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}$

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

$r=\sqrt{\frac{1}{4\pi {ϵ}_0}\frac{q}{E}}$

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

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

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

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

Hence, the distance from x can be calculated as:

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

And, the distance from y can be calculated as:

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

Therefore, the coordinates can be calculated as:

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

Part (b) Step 3: Final answer

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

Part (c) Step 1: Given Information

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

Given electric field $=\overrightarrow{E}=(21,600\hat{i}-28,800\hat{j})N/C$

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:

Part (c) Step 2: Calculation

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:

$E=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}$

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

$r=\sqrt{\frac{1}{4\pi {ϵ}_0}\frac{q}{E}}$

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

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

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

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

Hence, the distance from x can be calculated as:

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

And, the distance from y can be calculated as:

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

Therefore, the coordinates can be calculated as:

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

Part (c) Step 3: Final answer

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