Question

The variation of electric field between two charges $q_1$ and $q_2$ along the line joining the charges is plotted against the distance $r$ from $q_1$ (taking the direction from $q_1$ to $q_2$ as positive) as shown in the figure. Find the correct statement among the following is

• A. $q_1$ and $q_2$ are positive and $|q_1|<|q_2|$
• B. $q_1$ and $q_2$ are positive and $|q_1|>|q_2|$
• C. $q_1$ is positive and $q_2$ is negative $|q_1|<|q_2|$
• D. $q_1$ and $q_2$ are negative and $|q_1|<|q_2|$

Answer 1

The correct option is C $q_1$ is positive and $q_2$ is negative $|q_1|<|q_2|$

Just to the right of $q_1,\overrightarrow{E}$ is positive, so we can say that $q_1$ is positive.

Just to the left of $q_2,\overrightarrow{E}$ is negative, so we can say that $q_2$ is negative.

Let the distance between charges $q_1$ and $q_2$ be $d$ and let $x$ be the distance from $q_1$ where net electric field is equal to zero.

$\therefore \frac{k{q}_1}{x^2}=\frac{k{q}_2}{(d-x{)}^2}$

$\Rightarrow \left|\frac{q_2}{q_1}\right|=\frac{(d-x{)}^2}{x^2}$

Further, as ${\overrightarrow{E}}_{\text{net}}=0$ is close to $q_1$, we can conclude that the magnitude of charge $q_2$ must be greater than the magnitude of charge $q_1$.

So, $|q_2|>|q_1|$

Hence, option (c) is the correct answer.