Question

Two free positive charges $4q$ and $q$ are at a distance $l$ apart. What charge $Q$ is needed to achieve equilibrium for the entire system and where should it be placed from the charge $q$?

Answer 1

Step 1: Given information

Two free positive charges $4q$and $q$ are at a distance $\mathit{l}$ apart.

Step 2: To find

We have to find the charge $Q$ is needed to achieve equilibrium for the entire system and where should it be placed from charge $q$.

Step 3: Calculation

According to Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the two charges considered to be point charges.

$F=k\frac{q_1{q}_2}{r^2}$

Where, $k$ is the constant of proportionality, $q_1$and $q_2$ are the point charges, and $\mathit{r}$ is the distance between the two point charges

Let $Q$ is at a distance $r$ from $q$ charge, and therefore $(l-r)$ from $4q$ charge.

So, for equilibrium, force between $q$ and $Q$ should be equal to force between $4q$ and$Q$ i.e, $k\frac{qQ}{r^2}=k\frac{4qQ}{{\left(l-r\right)}^2}$

$$\begin{gathered} k\frac{qQ}{r^2}=k\frac{4qQ}{{(l-r)}^2} \\ \frac{1}{r^2}=\frac{4}{{(l-r)}^2} \\ \Rightarrow (l-r{)}^2=4r^2 \\ {\left(l-r\right)}^2={\left(2r\right)}^2 \\ l-r=2r \\ 3r=l \\ r=\frac{l}{3} \end{gathered}$$

Thus, the charge $Q$ should be placed at a distance $\frac{l}{3}$​ from charge $q$. Now, applying the condition of equilibrium on $q$ charge.

$$\begin{gathered} \therefore k\frac{qQ}{{\left(r\right)}^2}\text{=}k\frac{4qq}{l^2} \\ \frac{Q}{{\left(r\right)}^2}=\frac{4q}{l^2} \\ \frac{Q}{{\left({\displaystyle \frac{l}{3}}\right)}^2}=\frac{4q}{l^2}\left[\because r=\frac{l}{3}\right] \\ \frac{Q}{{\displaystyle \frac{l^2}{9}}}=\frac{4q}{l^2} \\ q=\frac{4q}{l^2}\times \frac{l^2}{9} \\ Q=\frac{4q}{9} \end{gathered}$$

$4q$ and $q$are given to be free positive charges placed at a distance $l$ apart. So, charge $Q$ needed to achieve equilibrium for the entire system should be of negative sign.

Therefore, the needed charge is $\mathrm{Q}=-\frac{4\mathrm{q}}{9}$.