Question

Reproduce FIGURE Q23.2on your paper. For each part, draw a dot or dots on the figure to show any position or positions (other than infinity) where $\underset{E}{\to }$$=$$\underset{0}{\to }$.

Short answer

The position at which $\underset{E}{\to }$$=$$\underset{0}{\to }$is

Step-by-step solution

Given information (part a)

Consider the given figure

Q₁=4e

Q₂=1e

Where e=charge

Explanation (part a)

The position at which $\overrightarrow{E}=\overrightarrow{0}\text{is}\left|\overrightarrow{F_1}\right|=\left|\overrightarrow{F_2}\right|\dots \dots \left(1\right)$

Here,

$\underset{F1}{\to =}$Columbic force between the charge Q1 & test charge Q

localid="1648533539776" $\underset{F1}{\to }=$Columbic force between the charge Q2 & test charge Q

$\left|\overrightarrow{F}\right|=\text{Columbic force}=\frac{q_1{q}_2}{4\pi {\epsilon }_0{d}^2}\dots \dots \left(2\right)$

Where, q₁&q₂ are two charges separated by distance d.

It is to be noted that, choosing the position of the test charge should satisfy the condition that $\underset{F1}{\to }$$\&$$\underset{F2}{\to }$should be in opposite direction.

Now substituting the given data in eqns (1) & (2)

$$\begin{gathered} \Rightarrow \left|\frac{\left(4e\right)\left(Q\right)}{4\pi {\epsilon }_0{D}^2}\right|=\left|\frac{\left(1e\right)\left(Q\right)}{4\pi {\epsilon }_0{r}^2}\right| \\ \Rightarrow 2r=D\dots \dots \left(3\right) \end{gathered}$$

By considering eqn(3), for the present scenario, the final figure pertinent to the position at which $\underset{E}{\to }$$=$$\underset{0}{\to }$is as shown below

Given information (part b)

Consider the given figure

Q₁=4e

Q₂=1e

Where e=charge

Explanation (part b)

The position at which $\overrightarrow{E}=\overrightarrow{0}\text{is}\left|\overrightarrow{F_1}\right|=\left|\overrightarrow{F_2}\right|\dots \dots \left(1\right)$

It is to be noted that, choosing the position of the test charge should satisfy the condition that $\underset{F1}{\to }$$\&$$\underset{F2}{\to }$should be in opposite direction.

$\left|\overrightarrow{F}\right|=\text{Columbic force}=\frac{q_1{q}_2}{4\pi {\epsilon }_0{d}^2}\dots \dots \left(2\right)$

Now substituting the given data in eqns (1) $ (2)

$$\begin{gathered} \Rightarrow \left|\frac{\left(4e\right)\left(Q\right)}{4\pi {\epsilon }_0{D}^2}\right|=\left|\frac{(-1e)\left(Q\right)}{4\pi {\epsilon }_0{r}^2}\right| \\ \Rightarrow 2r=D\dots \dots \left(3\right) \end{gathered}$$

By considering eqn(3), for the present scenario, to the final figure pertinent to the position at which $\underset{E}{\to }$$=$$\underset{0}{\to }$is as shown below