Question

Two point charges $Q_1$ and $Q_2$ are positioned at points $1$ and $2$ respectively. The field intensity to the right of the charge $Q_2$ on the line that passes through the two charges varies according to a law that is represented schematically in the figure. The field intensity is assumed to be positive, if its direction coincides with the positive direction on the $x-$ axis. The distance between the charges is $l$. Then,

• A. $Q_1,Q_2=$ Negative
• B. $Q_1,Q_2=$ Positive
• C. $Q_1=$ Negative, $Q_2=$ Positive
• D. $Q_1=$ Positive, $Q_2=$ Negative

Answer 1

The correct option is D $Q_1=$ Positive, $Q_2=$ Negative

Between $x=l$ to $x=l+a$, the field intensity is negative. This means $Q_2$ at point $2$ is negative.
But for $x>l+a$, the field intensity is positive with maximum at $x=l+b$. This means $Q_1$ is positive and due to which net electric field at $x>l+a$ becomes positive.
Therefore, option (c) is correct.